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
A block of mass 3kg starts from rest and slides down a surface, which corresponds to a quarter of a circle of 1.6m 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 4m/s, what is the energy dissipated by friction as it slides down?
c) After the block reaches the horizontal surface with a speed of 4m/s, it stops after travelling a distance of 3m from the bottom. find the frictional force acting on the horizontal surface due to the block.
1 answer
1 answer