Law of Conservation of Mechanical Energy
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A body of mass falls freely from a height of on a platform of mass which is mounted on a spring having spring constant . The body sticks to the platform and the springs maximum compression is found to be. Given that , the value of will be close to:
A small bob tied at one end of a thin string of length is describing a vertical circle so that the maximum and minimum tension in the string are in the ratio. The velocity of the bob at the highest position is ______ . (take)
A ball is dropped from rest from a height of 12 m. If the ball loses 25% of its K.E. on striking the ground, what is the height to which it bounces? How do you account for the loss in K.E.
- x√3k2m
- x√2k3m
- x√2km
- x√3km
- tan−1(12)
- tan−1(2)
- sin−1(23)
- The block never reaches point B
- 6 m
- 9.86 m
- 8.33 m
- 6.23 m
- 2g(d+v202g)
- 12g(2gd+v20)
- v2o2g
- 2gd
Does a Compressed Spring have more Mass?
A stationary football is kicked by a player such that it attains a speed of just after the kick. Find the work done by the player on the ball if the mass of the ball is .
A person is pulling a mass m from ground on a rough hemispherical fixed surface up to the top of the hemisphere with the help of a light inextensible string as shown in the figure. The radius of the hemisphere is R. Find the work done by the tension in the string. Assume that the mass is moved with a negligible constant velocity, u. (The coefficient of friction is μ)
(R−1)Mgμ
(1+R)Mgμ
mgR(1−μ)
MgR(1+μ)
The work done in raising a mass of 15 gm from the ground to a table of 1m height is
15 J
152 J
1500 J
0.15 J
A block of mass M slides along the sides of a bowl as shown in the figure. The walls of the bowl are frictionless and the base has coefficient of friction 0.2. If the block is released from the top of the side, which is 1.5 m high, where will the block come to rest ? Given that the length of the base is 15 m
1 m from P
Mid point
2 m from P
At Q
- l
- 3l2
- 2l
- Tension in the string will never be zero.
- vB>vC>vD
- vB<vC<vD
- vB>vC<vD
- vB=vC=vD
A child builds a tower from three blocks. The blocks are uniform cubes of side . The blocks are initially all lying on the same horizontal surface and each block has a mass of .the work done by the child is
(Take g=10 m/s2)
- 2 m
- 3 m
- 5 m
- 4 m
- θ=cos−1(1925)
- θ=cos−1(1125)
- θ=cos−1(1325)
- θ=cos−1(1725)
If W1, W2 and W3 represent the work done in moving a particle from A to B along three different paths 1, 2 and 3 respectively (as shown) in the gravitational field of a point mass m, find the correct relation
W1 > W2 > W3
W1=W2=W3
W1 < W2 < W3
W2 > W1 > W3
- Mgl2
- √3Mgl2
- Mgl√2
- 2Mgl√3
(Take g=10 m/s2)
- 10 m/s
- 8.9 m/s
- 0.89 m/s
- 9 m/s
- Mgl2
- √3Mgl2
- Mgl√2
- 2Mgl√3
- 49 mgl
- 1318 mgl
- Zero
- 1118 mgl
- θ=π4
- π4 <θ <π2
- π2 <θ <3π4
- 3π4 <θ <π
- √gl m/s
- √7gl m/s
- √18gl m/s
- √9gl m/s
- √2 ms−1
- 2√2 ms−1
- 2 ms−1
- 2√3 ms−1
- 75D
- 32D
- 54D
- D
- 5√2m/s
- 10m/s
- 10√2m/s
- 20√2m/s
- √2gR
- √gR
- √2gR(1−2π)
- √2gR(2−π)
- mgh = 12kx2
- mg(h+x) = 12kx2
- mgh = 12k(x+h)2
- mg(h+x) = 12k(x+h)2
- v=⌊gl(2+√3)⌋1/2
- v=⌊2gl(2+√3)⌋1/2
- v=⌊gl(1+√3)⌋1/2
- v=⌊gl(2+√3)⌋1/3