# 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 m/s^−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 m/s^−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.

1
by puja147

2015-06-01T12:33:25+05:30

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M = 3 kg      R = 1.6 meters

We assume that  the block slides down the circular arc without jumping off..

1)  The potential energy of the block is converted into the kinetic energy.
1/2 m v² = m g R
=> v² = 2 g R = 2 * 10 * 1.6
=> v = 5.6 m/sec

2)  v = 4 m/s,  at the bottom...  so energy lost or dissipated due to heat
mg R - 1/2 m v² = 3 * 9.81 * 1.6 - 1/2 * 3 * 4²
=  23.088  Joules

3) kinetic energy in the block when it starts on the horizontal surface
= 1/2  m v² = 1/2 * 3 * 4² = 24 Joules
frictional force * distance travelled = work done by friction force
24 joules = Ff * 3 m  =>  friction = 8 Newtons

we can also calculate the coefficient of friction along the surface :
friction force / m g = 8 / (3 * 9.81) = 0.271