Energy is always conserved, but to find out where it has gone we sometimes need to look inside matter. Macroscopic processes can lead to microscopic changes.
When work is done on a block on a slope it gains both kinetic and potential energy but when the force is removed the kinetic energy is converted to potential energy. At the top, the potential energy gained is equal to the total work that was done:
\(Fd = mgh + {1\over2}mv^2\)
If the ball itself could squash, the energy could be considered to have transferred to elastic potential energy.Use quizzes to practise application of theory.
START QUIZ!
A constant force of 100 N is applied to a block as shown.
If the frictional force is 10 N calculate
The work done on the block by the person pulling it = J
The work done against friction = J
The increase in PE when the block is at the top of the slope = J
The gain in KE when the block is at the top = J
Work = 5 x 100 = 500 J
Work against friction = 5 x 10 = 50 J
gain in PE = 10 x 10 x 2 = 200 J
Work done = gain in PE + KE + work against friction, KE = 500 - 200 - 50 = 250 J
A constant force of 100 N is applied to a block as shown.
If there is no friction calculate
The work done on the block by the person pulling it = J
The increase in PE when the block is at the top of the slope = J
The gain in KE when the block is at the top = J
Work = 5 x 100 = 500 J
gain in PE = 10 x 10 x 2 = 200 J
Work done = gain in PE + KE KE = 500 - 200 = 300 J
A block is sent up the hill shown with a velocity v.
The block just reaches the top of the hill and falls off the other side
The frictional force is 60 N.
The statement that the block "just reaches the top" implies that the velocity at the top = ms-1
The work done against friction = J
The gain in PE = J
The loss in KE = J
The original KE = J
The original velocity = ms-1
Work done against friction = 60 x 5 = 300 J
Gain in PE = 10 x 10 x 2 = 200 J
Loss in KE = 500 J
1/2mv2 = 500 = 5 v2
A block travelling at 10 ms-1 travels up the hill shown arriving at the top with a velocity of 4 ms-1
Calculate
The KE of the block at the bottom = J
The KE at the top = J
The PE at the top = J
The work done against friction = J
The frictional force = N
1/2 x 10 x 102 = 500 J
1/2 x 10 x 42 = 80 J
10 x 10 x 2 = 200 J
Loss of KE = gain in PE + work against friction
Work against friction = 500 -80 -200 = 220 J
Force = work / dist = 220 / 5 = 44 N
A ball drops from a height of 10 m as shown
When the ball bounces its velocity is reduced to half what it was.
Calculate
The KE just before hitting the ground = J
The KE just after the bounce = J
The energy lost in the form of thermal energy = J
The height reached after the bounce = m
gain in KE = loss of PE = mgh = 5 x 10 x 10 = 500 J
If velocity is 1/2 the KE is 1/4 = 125 J
mgh = 125 = 2.5 m
How much of Conservation of energy have you understood?
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