A man of mass 50 kg climbs up a ladder of height 10m. Calculate: (i) the work done by the man, (ii) the increase in his potential energy. (g= 9.8m s-2). Solutions: Given. Mass of man = 50 kg. Height of ladder, h 2 = 10 m (i) Work done by man = mgh 2 = 50 × 9.8 × 10 = 4900 J (ii) Increase in his potential energy. Height, h 2 = 10 m. Reference. Kinetic Energy is the energy an object has owing to its motion. In classical mechanics, kinetic energy (KE) is equal to half of an object's mass (1/2*m) multiplied by the velocity squared. For example, if a an object with a mass of 10 kg (m = 10 kg) is moving at a velocity of 5 meters per second (v = 5 m/s), the kinetic energy is equal to 125. In this example, a 3 kilogram mass, at a height of 5 meters, while acted on by Earth's gravity would have 147.15 Joules of potential energy, PE = 3kg * 9.81 m/s 2 * 5m = 147.15 J. 9.81 meters per second squared (or more accurately 9.80665 m/s 2 ) is widely accepted among scientists as a working average value for Earth's gravitational pull

A body of mass 5 k g is taken from a height of 5 m to 1 0 m. Find the increase in its potential energy. Find the increase in its potential energy. Take g = 1 0 m s − 2 PE grav. = m * h * g. Where: m - mass; h - height; g - the gravitational field strength (9.81 on Earth) The formula is relatively simple. An object which is not raised above the ground will have a height of zero and therefore zero potential energy. When you double the mass or the height of an object, its potential energy will also double

Formula. The calculation formula used for this tool is: F g = m · g. Symbols. F g = weight or force due to gravity of object. m = mass of object. g = local gravity (e.g. standard earth gravity or g 0 = 9.80665ms -2) Mass of Object. Enter the mass of the object that you wish to determine the gravitational weight As the kinetic energy of body moving with velocity of 10 m/s is 500 j So equating mgh =500j h=500/mg H=500/10(9.8) H=5.1

- m = 10 kg h = 5 m g = 10 m/s² Mechanical Energy = Potential Energy + Kinetic Energy Potential Energy at h of 5 m = mgh = (10) ×(10) × (5) = 500 J. i) Energy neither can be created or destroyed but can be converter in to another form. as its a Frictionless surface so no energy loss in any other form. so. Total Mechanical Energy at h of 2m = 500
- At what height above the ground must a body of mass 10kg be situated in order to have potential energy in value to the kinetic energy possessed by another body of mass 10kg moving with a velocity of 10m-s . physics. The two masses in the Atwood's machine shown in the figure are initially at rest at the same height
- ANSWER:-Here m = 5kg; h=10m ; g=10m/s^2 Potential energy of the object, E p = mgh =5 x 10 x 10 = 500J. Hence answer is 500
- Estimate the total mass of the planet in kilograms. Estimate the radius of the planet, from its center to its surface, in meters. Divide the total mass by radius squared. Multiply the result by the universal Gravitational constant, 6.67×10-11 N·m 2 ·kg-2. The result is the gravitational force of the planet, which is also its free fall.
- v = a t = 3.11 m/s2 1.46 s = 4.53 m/s c. What will be the kinetic energy of the cart when it reaches the bottom of the incline? Ans. 2TKE = ½mv2 = ½(2.50 kg)(4.53 m/s) = 25.66 J Gravitational Potential Energy GPE = mg h 12. A 5.0 kg mass is initially sitting on the floor when it is lifted onto a table 1.15 meters high at a constant speed. a
- A boy of mass 50 kg runs up a staircase of 45 steps in 9 s. If the height of each step is 15 cm, find his power. Take g = 10 m s-2 NCERT Class IX Sciences - Main Course Book Chapter 11. Work and Energ

* A crane can lift 5,000 kg of mass to the height of 10 m in 10 seconds*. What is its power? Thanks for A2A I think you want to know the power of the crane, then its wrong term to use. Cranes are known by their capacity. Edison Ooi gave good mathemat.. A metal ball of mass 2 k g is allowed to fall freely from rest from a height of 5 m above the ground (Take g = 1 0 m s − 2). Calculate the potential energy possessed by the ball when initially at rest (a) (i) The total mechanical energy is the total potential energy it possesses before its fall. E = mgh = 10 × 10 × 5 = 500 J . It is equal to the maximum potential energy E = mgh = 10 × 10 × 5 = 500 J. (b) (i) The weight of the metre scale acts at its centre of gravity i.e., at 50 cm mark. By the concept of moments, we hav Mass = 100 kg. Height = 5 m. g = 9.8 m/s 2. PE = mgh = 4900J. Work done = potential energy. Q20. State whether the following statement is true or false: The potential energy of a body of mass 1 kg kept at a height of 1m is 1J. Answer: False. PE = mgh = (1) (9.8) (1) = 9.8J. Q21. What happens to the potential energy of a body when its height is.

A man of mass 50 kg jumps to a height of 1 m. His potential energy at the highest point is (g = 10 m/s 2): A. 50 J. B. 500 J. C. 5 J. D. An iron sphere of mass 10 kg has the same diameter as an aluminum sphere of mass 3.5 kg. Both spheres are dropped simultaneously from a tower. When they are 10 m above the ground, they have the same (g= 9.8 m/s²) Answer: a) Potential energy, PE=mgh b) m= 80 kg, g = 9.8 m/s², h= 50 m Total energy = mgh = 80 x 9.8 x 50=39200J. Question 20. a) State work energy theorem. (MAY-2016) b) Show that the potential energy of a body is . completely converted into kinetic energy during its free fall under the gravity. c) A man carefully brings down a. * A bullet of mass 50 g is fired from below into a suspended object of mass 450 g*. The object rises through a

CONCEPT:. Potential energy: The energy of an object due to its position is called potential energy.It is denoted by PE. Mathematically potential energy can be written as P.E of object = m g h Where m = mass of an object, g = acceleration due to gravity and h = height The equation of motion is given by:. S = u t + (1/2) a t 2. Where S is displacement, u is initial velocity, a is acceleration. At what height above the ground must a body of mass 10kg be situated inorder to have potential energy equal in value to the kinetic energy possessed by another body of mass 10kg moving with a velocity of 10m/s? Physics. Calculate the work done by a stone mason lifting a stone of mass 15kg through a height of 2m.Take g=10N/Kg Activity 11.15 An object of mass 20 kg is dropped from a height of 4 m. Fill in the blanks in the following table by computing the potential energy and kinetic energy in each case. Take g = 10 m/s2Mass of the object = m = 20 kgAcceleration due to gravity = g = 10 m/s2At Height = 4

- Derive an expression for potential energy of an object of mass 'm' that has been raised to a height 'h' from the ground. A body of mass 20 kg is lifted up by 10 meters. Calculate its potential energy. If this body is allowed to fall, find its kinetic energy just before it touches the ground. (Take g = 10m/s 2) SECTION-B. Question 22
- 2. × 30m g = (10 m/s. 2) = 24000J = 24 KJ . Potential Energy (P.E) is the energy associated with the position of a body relative to the earth's surface. For example, lifted masses above the earth's surface possess potential energy. The term potential means stored. The potential energy of a body of mass (m) lifted to a height of 'h' abov
- Take g = 10 m s-2 Solution 5: Mass = 0.5 kg Energy = 1 J Gravitational potential energy = mgh 1 = 0.5 × 10 × h 1 = 5h Height, h = 0.2 m Question 6: A boy weighing 25 kgf climbs up from the first floor at height 3 m above the ground to the third floor at height 9m above the ground. What will be the increase in his gravitational potential energy
- At Height = 5 m Mass of the object = 40 kg Height of the object = 5 m Acceleration due to gravity = g = 10 m/s2 Potential Energy = mgh = 40 × 10 × 5 = 2000 J Object falls half way Object falls half way from height 5 m to 2.5 m We need to find its Kinetic Energy First, lets find velocity When ball drops, Initial velocity = u = 0 m/s Distance.
- Free online Kinetic Energy calculator with which you can calculate the energy of an object or body in motion given its mass and velocity. The calculator can be used to solve for mass or velocity given the other two. Supports multiple metrics like meters per second (m/s), km per hour, miles per hour, yards and feet per second. Mass units in metric and imperial units

v = a **t** = 3.11 m/s2 1.46 s = 4.53 m/s c. What will be the kinetic energy of the cart when it reaches the bottom of the incline? Ans. 2TKE = ½mv2 = ½(2.50 kg)(4.53 m/s) = 25.66 J Gravitational **Potential** Energy GPE = mg h 12. A **5.0** **kg** **mass** **is** initially sitting on the floor when it is lifted onto a table 1.15 meters high at a constant speed. **a** v2 r smg = m v2 r s = v2 gr = (50:2 10 2)2 9:81 0:302 = 0:085 0.9 A hawk ies in a horizontal arc of radius 14.2 m at a constant speed 4.10 m/s. • a) Find its centripetal acceleration. • b) It continues to y along the same horizontal arc, but increases its speed at the rate of 1.00 m/s2. Find the acceleration in this situation at the moment.

- Is the kinetic energy of the system changing? Explain why or why not. 10(b) A bullet of mass 2.0 g is moving horizontally with a speed of 500.0m/s. The bullet strikes and becomes embedded in the edge of a solid disk of mass M=3.2 kg and radius R=0.5m. The disk is free to rotate around its axis and is initially at rest
- Chapter 10: Rotational Kinematics and Energy thJames S. Walker, Physics, 4 Edition 10.75) A 2.0 kg solid cylinder (radius = 0.10 m, length = 0.60 m) is released from rest at the top of a ramp and allowed to roll without slipping. The ramp is 0.75 m high and 5.0 m long. A) When the cylinder reaches the bottom of the ramp, what is its total kinetic energy
- A bullet of mass 50 g is fired from below into a suspended object of mass 450 g. The object rises through a height of 1.8 m with bullet remaining inside the object. Find the speed of the bullet. Take g = 10 ms-2. Solution. m 1 = 50 g = 0.05 kg; m 2 = 450 g = 0.45kg. The speed of the bullet is u 1. The second body is at rest u 2 = 0 . Let the.
- First, we need to convert km/h to m/s, which gives us 27 / 3.6 = 7.5 m/s. Then we apply the first equation since we know the deformation distance, which is 75 cm = 0.75 meters. Replacing in the formula we get F avg = 0.5 · 2400 · 7.5 2 / 0.75 = 90 kN and a maximum impact force of 180 kN. ( calculation link) Example 2: Using the situation in.
- 14. The total length of a Hooke's law spring with a mass m = 0.2 kg hung under is 0.2 m. The total length of the same spring with a mass of 0.7kg hanging under is 0.25 m. a) Find the spring constant k of this spring. b) How much elastic potential energy is stored in the spring when the mass hung under is m = 0.7 kg? 15. What major.

- Question 19: A body of mass 0.2 kg falls from a height of 10 m to a height of 6 m above the ground. Find the loss in potential energy taking place in the body. [g = 10ms-2] Solution: Question 20: A ball of mass 200 g falls from a height of 5 m. What will be its kinetic energy when it just reaches the ground? (g = 9.8 m s-2) Solution
- A skier with a mass of 88 kg hits a ramp of snow and becomes airborne. At the highest point of flight, the skier is 13 m above the ground. The skier lands moving vertically at 16 m/s. What is the skier's kinetic energy as he lands? a. 7.0 x 10^2J b. 11000J c. 5500J d. 18,000J KE=0.5m(v)^
- Gravitational Potential Energy Calculator. Gravitational Potential Energy (GPE) is defined as the energy possesed by an object due to its position in the gravitational field. GPE is most commonly used for an object near the surface of the Earth. The gravitational acceleration is assumed to be a constant whose values is equal to 9.8 m/s 2

** g + (h 1−h 2) µ K =6**.0 m 6. A 0.5 kg ball moving horizontally at 10.0 m/s strikes a vertical wall and rebounds in the opposite direction at 10.0 m/s. If the collision took place in 0.1 s, what was the magnitude of the average force on the wall? (1) 100 N (2) 200 N (3) 0 N (4) 50 N (5) 10 N J=Δp=F avΔt m(v 2−v 1) Δt = (0.5kg)(10−()m/s. 2 10 9:81 6 = 0:425 0.7 A block of mass m 1 = 18:0 kg is connected to a block of mass m 2 = 32.0 kg by a massless string that passes over a light, frictionless pulley. The 32.0-kg block is connected to a spring that has negligible mass and a force constant of k = 220 N/m as shown in the gure below

GPE (joules) = mass (kg) x gravitational acceleration (9.8 m/s/s) x height (m) GPE = m g h A second form of potential energy is elastic potential energy. Elastic potential energy is the energy stored in elastic materials as the result of their stretching or compressing Gravitational potential energy and work done. A ball of mass 0.4 kg is lifted to a height of 2.5 m. the 25 kg mass and the 50 kg would fall at the same rate of 10 m/s 2 Question Number. 13. A force of 15N is needed to move a body of mass 30kg along a footpath with uniform velocity. Find the coefficient of dynamic friction.(take g as 10m/s/s). Option A. 1/20. Option B. 1/2. Option C. 20 The formula for kinetic energy is: KE = 1/2 x mass x velocity squared. We can plug in the information for the ball into this equation to determine the kinetic energy. The mass is 0.5 kg and the velocity is 10 meters per second. So: KE = (1/2)(0.5)(10)^2. Thus the kinetic energy of the ball is 25 J if body is on the ground h= then, h' = .m 5. Q. A rocket of mass 3x10^6 kg taken from a launching pad and acquires vertical velocity of 1 km/s and an altitude if 25m. Calculate its Potential Energy and Kinetic Energy. Ans: Given: m = 3x10^6 kg, v = 1 km/s or 1000m/s and h = 25m. Potential Energy = mgh = 3x10^6 kg x 10m/s2 x 25m =7.5x108

The force of gravity, g = 9.8 m/s 2 Gravity accelerates you at 9.8 meters per second per second. After one second, you're falling 9.8 m/s. After two seconds, you're falling 19.6 m/s, and so on. Time to splat: sqrt ( 2 * height / 9.8 ) It's the square root because you fall faster the longer you fall A satellite of mass 100 kg performs uniform circular motion around the earth at a height of 6400 km from the earth's surface. Find its gravitational potential energy. [g = 9.8 m/s 2, R = 6400 km] Solution: Data: m = 100 kg, g = 9.8 m/s 2, R = 6400 km = 6.4 × 10 6 m, h = 6.4 × 10 6 m The gravitational potential energy of the satellite. ** 15**. A 2 kg mass has 40 J of potential energy with respect to the ground. Approximately how far is it located above the ground? P = 100 J / 50 s. P = 2 W. D) 4.0 W . 5. An object is raised above the ground gaining a certain amount of potential energy. If the same object is raised twice as high it gains PE = m g h. PE = (2 kg) (10 m/s 2.

- The elastic potential energy is stored in the spring is 2 J. What is the spring constant if the elongation of the spring is 10 cm? A. 400 N/m B. 300 N/m C. 200 N/m D. 100 N/m E. 50 N/m 21. A heavy block is suspended from a vertical spring. The elastic potential energy is stored in the spring is 0.8 J
- When it is dropped from the height initial velocity u = 0 m/s. according to equation. v 2-u 2 =2gh. v 2-0=2x10x50x10-2. K.E. = ½ mv 2 = ½ x 10x10 = 50J. when mass will hit the ground it will have same kinetic energy as that was potential energy before dropping. This is also in agreement with law of conservation of energy which states that.
- 41 When the body is allowed to fall, its potential energy reduces as it approaches the ground. Potential energy = Work done = F x d but F = mg and d = h P.E = mgh Where g = 10m/s 2 and h is the height above the ground. Example1; A stone of mass 8kg is lifted through a height of 2metres.Find the potential energy the stone develops (Take g = 10m/s 2) P.E = mgh = 8 x 10 x 2 = 160J Question2; A.
- Calculate: (i) the loss in potential energy of the body, (ii) the total energy possessed by the body at any instant? (Take g = 10 ms-2). Answer 4. Question 5. Calculate the height through which a body of mass 0.5 kg should be lifted if the energy spent in doing so is 1.0 J. Take g = 10m/s - 2. Answer 5. Mass = 0.5 kg. Energy = 1
- MCQ Questions for Class 9 Science: Ch 11 Work and Energy. 1. Work done by centripetal force is. 2. When a body falls freely towards the earth, then its total energy. 3. If work is done at a faster rate, then. 4. Moon revolves around the earth due to gravitational force (F) of earth on moon

* Q*.2 Define energy and define its SI unit. Sol. Energy is the capacity to do work. The unit of energy of an object is joule. 1 Joule = 1N × 1m.* Q*.3 An archer stretches a bow to release an arrow to hit the target at a distance of 10 m. Explain who does the work,in which form is the energy possessed the bow and the arrow The KE of any object can be computed if the mass (m) and speed (v) are known. Simply use the equation. KE=0.5*m*v 2. In this case, the 10-N object has a mass of approximately 1 kg (use F grav = m*g). The speed is 1 m/s. Now plug and chug to yield KE of approximately 0.5 J

GPE = mass * g * height. GPE = 2kg * 9.8 m/s 2 * 10m. GPE = 196 J. Potential Energy and Work. The potential energy is equal to the amount of work done to get an object into its position. For example, if you were to lift a book off the floor and place it on a table. The potential energy of the book on the table will equal the amount of work it. Certain force acting on a 20 kg mass ch.urlyc its velocity from 5 m/s to 2 m/s. Calcukate the work done by the Jbrce. A body at a particular height has maximum potential energy, when it falls the potential energy is transformed into kinetic energy, when the object is about to touch the ground, the body has maximum velocity and hence the. object, but later on the potential energy can be released. The most familiar example of potential energy is the gravitational po-tential energy, when the work must be done on an object to lift it up. For example, if on object has mass m and is lifted from height x 1 to height x 2, then the work done by gravitational force is W grav =! −mgˆi.

CONCEPT:. Potential energy (PE): The energy of an object due to its position is called potential energy. Mathematically potential energy can be written as . P.E of object = m g h Where m = mass of an object, g = acceleration due to gravity, and h = height Work is said to be done by a force when the body is displaced actually through some distance in the direction of the applied force 14c, kinetic energy = 1/2 mv 2 18 = 1/2 x 4v 2 v = 3 m/s. 15. An empty lift is counterbalanced by a heavy piece of metal. Some people of combined mass 350 kg enter the lift and operate it. The lift rises 50 m in 60 s. Calculate a. the work done in raising the people b. the power required to do this (take weight of 1kg to be 10N) Solutio 2. A 2.00 -kg object traveling east at 20.0 m/s collides with a 3.00 -kg object traveling west at 10.0 m/s. After the collision, the 2.00 -kg object has a velocity 5.00 m/s to the west. How much kinetic energy was lost during the collision? Solution: We don't know the nal velocity of the second object after the collision, we'll need to nd it (A) 10 m/s 2 (B) 5 m/s 2 (C) 15 m/s 2 (D) 3 m/s 2 95. A spherical ball of mass 20 kg is stationary at the top of a hill of height 100 m. It rolls down a smooth surface to the ground, then climbs up another hill of height 30 m and finally rolls down to a horizontal base at a height of 20 m above the ground. The velocity attained by the ball is.

An object of mass m when raised To height h possess a potential energy of 1200 J Find the new potential energy if the same object is raised to height - Tutorix A body possess potential energy of $460\ J$ whose mass is $20\ kg$ and is raised to a certain height. its velocity is doubled ? Asked on 15th Jul, 2021. Find the energy possessed. An engine can pump 30,000 litres of water to a vertical height of 45 metres in 10 minutes. Calculate the work done by the machine and its power. [g = 9.8 m/s 2; Density of water = 10 3 kg/m 3, 1000 litre = 1 m 3]. Solution: Volume of water raised = 30,000 litres = \(\frac{30,000}{1000}\)m 3 = 30 m 3. Mass of water raised, M = Volume × Densit (Take g = 10 m s-2 and 1 hp = 750 watts). E-3. A labourer lifts 100 stones to a height of 6 metre in two minute. If mass of each stone be one kilogram, calculate the average power. Given : g = 10 m s-2. E-4. The power of a pump motor is 2 kilowatt. How much water per minute can it raise to a height of 10 metre? Given : g = 10 m s-2. E-5 Let us calculate the work done in lifting an object of mass m through a height h, such as in Figure 1.If the object is lifted straight up at constant speed, then the force needed to lift it is equal to its weight mg.The work done on the mass is then W = Fd = mgh.We define this to be the gravitational potential energy (PE g) put into (or gained by) the object-Earth system Sol. (a) Energy possessed by a body due to its motion is called its kinetic energy.Kinetic energy is directly proportional to the mass and square of the velocity of the moving body. K.E= 1/2mv2 (b) When a body is thrown vertically upwards against the force of gravity, its Kinetic energy goes on decreasing as its velocity decreases

E o = m o c 2 . E o = (1.672 x 10 - 27 kg)(3 x 10 8 m/s) 2 . E o = 5.016 x 10 - 11 J. This is perfectly okay as it is. However, instead of dealing with such small numbers, it is customary to talk about these energies in units of MeV -- millions of electron-Volts. We will explore MeV's more in PHY 1360 (next semester) A tank is full of water. Find the work W required to pump the water out of the spout. (Use 9.8 m/s^2 for g. Use 1000 kg/m^3 as the weight density of water. Assume that a = 4 m, b = 4 m, c = 12 m, a..

19.6 m/s 2. If an object on the surface of this planet weighs 980. newtons, the mass of the object is A) 50.0 kg C) 490. N B) 100. kg D) 908 N 2. If a car is traveling at an average speed of 60 kilometers per hour, how long does it take to travel 12 kilometers? A) 0.2 hour C) 0.72 hour B) 0.5 hour D) 5.0 hours 3 3. (G9) A .50-kg mass at the end of a spring vibrates 3.0 times per second with an amplitude of 0.15 m. Determine: a. The velocity when it passes the equilibrium point. Answer: 2.8 m/s b. The velocity when it is 0.10 m from equilibrium. Answer: 2.1 m/s c. The total energy of the system. Answer: 2.0 J d. The equation describing the motion of. Gravitational Potential Energy In raising a mass m from y 1 to y 2 by a height h, the work done by the external force is The gravitational potential energy is raised by ' U G mg( y 2 y 1) mgh W G mg d mg( y 2 y 1) mgh & & The work done by the gravitational force Therefore, W G ' U G The work done by the gravitational force is equal to the loss.

The simplest formula is : GPE = m*g*h. Where : GPE = gravitational potential energy (joules j) m = mass in kilogram (kg) g = acceleration due to gravity (m/s²) h = change in height h rather than the usual Δh (Note that h is positive when the final height is greater than the initial height, and vice versa), in meters (m A crate having mass 50.0 kg falls horizontally off the back of the flatbed truck, which is traveling at 100 km/h. Find the value of the coefficient of kinetic friction between the road and crate if the crate slides 50 m on the road in coming to rest. The initial speed of the crate is the same as the truck, 100 km/h Chapter 5 • Dimensional Analysis and Similarity 5.1 For axial flow through a circular tube, the Reynolds number for transition to turbulence is approximately 2300 [see Eq. (6.2)], based upon the diameter and average velocity. If d = 5 cm and the fluid is kerosene at 20°C, find the volume flow rate in m3/h which causes transition. Solution: For kerosene at 20°C, take ρ = 804 kg/m3 and μ. Answer: 12.7 m/s. This problem is similar to the above. During the entire descent down the hill, gravity is doing work on the sledder and friction is doing negative work on the sledder. Friction is a non-conservative force and will alter the total mechanical energy of the sledder. The equation to be used is We have both m and g. m equals 100 kg, and g equals 9.8 m/s 2, because we're looking for the weight of the object on the surface of the earth. We set up our equation next: F = 100 kg x 9.8 m/s 2. This gives us the final answer. On the surface of the earth, an object with a mass of 100 kg will weigh approximately 980 Newtons. F = 980 N

A body of mass 10 kg is moving with a velocity 20 m s-1. If the mass of the body is doubled and its velocity is halved, find the ratio of the initial kinetic energy to the final kinetic energy. If the mass of the body is doubled and its velocity is halved, find the ratio of the initial kinetic energy to the final kinetic energy 1. (a) Justify that a body at a greater height has larger energy. (b) A body of mass 2 kg is thrown up at a velocity of 10 m/s. Find the kinetic energy of the body at the time of throw. Also, find the potential energy of the body at the highest point. The value of g = 10m/ (1)1 m/s (3)0.5 m/s (2)2 m/s (4)0 m/s 7. If the speed of a moving object is doubled, which quantity associated with the object must also double? (1)its momentum (2)its kinetic energy (3)its acceleration (4)its gravitational potential energy 8. A 2-newton force acts on a mass. If the momentum of the mass changes by 120 kg.-meters/sec., the force. velocity (υ) = 1 m/s g = 10 m/s 2 or, 100 = m x 10 x 1 ∴ m = 10 kg. Question 6. A car of mass 2000 kg is lifted up a distance of 30 m by a crane in 1 minute. A second crane does the same job in 2 minutes. What is the power applied by each crane? Answer: Given, mass of the car to be lifted, m = 2000 kg height through which the car is to be. TIME: 9:00 a.m. to 11:05 a.m. Answer ALL questions. You are requested to show your working and to write the units where necessary. When necessary, take g, acceleration due to gravity, as 10 m/s2. Density Pressure Moments Energy and Work \ Force and Motion Waves Electricity Electromagnetism

Illustration 8 : Hetasvi, having her own mass 50 kg, climbs 20 m height along with 30 kg mass in 40 s. Calculate her power and work done. (take g = 10m/s2) Solution : Total mass m = 50 + 30 = 80 kg, height h = 20 m, time t = 40 s Work done against gravitational force W = m g h = 80 × 10 × 20 = 16000 J Power P = W/t = 16000 J/40 s = 400 Using that definition, the gravitational potential energy of a system of masses m1 m 1 and M2 M 2 at a distance r r using gravitational constant G G is. U= −Gm1M2 r +K U = − G m 1 M 2 r + K. where K K is the constant of integration. Choosing the convention that K =0 K = 0 makes calculations simpler, albeit at the cost of making U U negative A skater with a mass of 52.0 kg moving at 2.5 m/s glides to a stop over a distance of 24.0 m. How much work did the friction of 2.10 m high. What is the potential energy of the book relative to the desk? PE! mg(h f h i)! tial energy stored in the rider's body. 17. A skier starts from rest at the top of Wile E began to fall from a higher height 20 seconds after he dropped the anvil from the weather balloon and passed the anvil 4 seconds after beginning his fall. The anvil should have traveled 1.75 miles down by the point he fell past it, calculated by your web page. [ Gravity g: m/s 2 ] Unit system: [10] 2021/04/30 07:46 Male / 50.

Shown below is a small particle of mass 20 g that is moving at a speed of 10.0 m/s when it collides and sticks to the edge of a uniform solid cylinder. The cylinder is free to rotate about its axis through its center and is perpendicular to the page. The cylinder has a mass of 0.5 kg and a radius of 10 cm, and is initially at rest a = 20 m/s 2. t = 10 s. v = u + at = 0 + 20 x 10 = 200 m/s (2) A mass of 10 kg is dropped from the top of a building which is 100 metres tall. How long does it take to reach the ground? In this example, in theory it doesn't make any difference what the value for the mass is, the acceleration due to gravity is g irrespective of the mass 1. Potential energy is the energy possessed by a body by virtue of its position. It is given by the following relation: P.E. = m g h Please note: All bodies or objects possess potential energy due to their position. 2. Kinetic energy is the energy possessed by a body by virtue of its motion. Example: A speeding train, flowing water, etc

Weight is given by the formula, W = mg. W is the weight of the object in newtons. m is the mass of the object in kilograms. g is the gravitational acceleration. On Earth, g = 9.8 m/s2. So, the object's weight on Earth will be: W = 50 kg ⋅ 9.8 m/s2. = 490 N mechanical energy per step per kg of body mass. A) If a 60 kg person develops a power of 70 Watt An object (0.2 kg) is dropped from a height of 35 m. potential energy: mgh g=9.81 m/s2 ME i -ME f = E nc (KE i+PE i) - (KE f+PE f)= E nc. MSU Physics 231 Fall 2015 44 Question An outfielder who is 2M tall throws a baseball of 0.15 kg at 14c, kinetic energy = 1/2 mv 2 18 = 1/2 x 4v 2 v = 3 m/s 15. An empty lift is counterbalanced by a heavy piece of metal. Some people of combined mass 350 kg enter the lift and operate it. The lift rises 50 m in 60 s. Calculate a. the work done in raising the people b. the power required to do this (take weight of 1kg to be 10N) Solutio 1. Calculate the work done by a stone mason lifting a stone of mass 15 kg through a height of 2.0 m. (take g=10N/kg) Solution. Work done = force × distance = (15× 10) × 2 = 300 Nm or 300 J. 2. A girl of mass 50 kg walks up a flight of 12 steps. If each step is 30 cm high, calculate the work done by the girl climbing the stairs. Solutio

Acceleration = m/s 2 compared to 9.8 m/s² for freefall. If the height of the incline is h= m, then the time to slide down the incline from rest would be t= seconds, compared to a time of t= seconds to drop from that height. The speed at the bottom of the incline would be m/s. These calculations can be done with the motion equations (Take g =10ms −2) 1. Find the kinetic energy of the following: (a) a runner of mass 50 kg running at 6 ms −1 (b) a hovercraft of mass 105 kg travelling at 25 ms −1 (c) an electron of mass 9×10 −28 g, speed 2×10 8 ms−1 (d) a car of mass 1300 kg travelling at 60 kph. 2. A boy of mass 20 kg rides a bicycle of mass 15kg along a road at.

Kinetic energy refers to the energy present in the object due to the motion of a body. Potential energy refers to the energy present in an object due to its position. 4. K.E = (1/2)mv 2 where m is mass and v is speed. P.E = mgh where m is mass, g is gravity and h is height constant speed of 10 m/s. At B, the car encounters an unbanked curve of radius 50 m. The car follows the road from B to C traveling at a constant speed of 10 m/s while the 10 m/s 2 (e) zero m/s 2 (b) 5 m/s 2 (d) 20 m/s 2 and a mass MM = 6.40 × 1023 kg. The mission of the satellite is to observe the Martia The kinetic energy of an object is dependent both the mass of the object and its velocity. K.E. = 1/2 m v 2. Thus a 3,000 lb car moving at 50 mph will transfer more kinetic energy than a 2,000 lb car moving at the same rate. Energy can be stored in an object by lifting it up. The amount of potential energy would b be r = 1.25 kg/m3. Analysis Kinetic energy is the only form of mechanical energy the wind possesses, and it can be converted to work entirely. Therefore, the power potential of the wind is its kinetic energy, which is V2/2 per unit mass, and m&V2/2 for a given mass flow rate: 0.050 kJ/kg 1000 m/s 1 kJ/kg 2 (10 m/s) 2 22 2 2 mech ˜= ¯ ˆ Á Ë.

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