A block of mass 3 kg starts from rest and slides down a surface, which corresponds to a quarter of a circle of 1.6 m radius. a) If the curved surface is smooth, calculate the speed of the block at the bottom. b) If the block’s speed at the bottom is 4 ms−1, what is the energy dissipated by friction as it slides down? c) After the block reaches the horizontal surface with a speed of 4 ms−1, it stops after travelling a distance of 3 m from the bottom. Find the frictional force acting on the horizontal surface due to the block.
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1)
change in KE = change in PE
1/2 m v² = m g R
v = √[2 g R] = √[2*10*1.6] = √32 m/sec
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2)
Loss of energy if the speed is only 4 m/s : 1/2 * 3 kg * [ (√32)² - 4² ]
= 24 Joules
This is the energy dissipated.
===========================
3)
Energy dissipated on the horizontal surface = 1/2 * 3 kg * 4² m²/s² = 24 Joules
work done by friction = Force * distance
24 J = Force * 3 m
Force = 8 N
change in KE = change in PE
1/2 m v² = m g R
v = √[2 g R] = √[2*10*1.6] = √32 m/sec
============
2)
Loss of energy if the speed is only 4 m/s : 1/2 * 3 kg * [ (√32)² - 4² ]
= 24 Joules
This is the energy dissipated.
===========================
3)
Energy dissipated on the horizontal surface = 1/2 * 3 kg * 4² m²/s² = 24 Joules
work done by friction = Force * distance
24 J = Force * 3 m
Force = 8 N
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