Question

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.

Answer 1

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