Answer to Question #104489 in Classical Mechanics for Neha Singh

Question

A black of mass 4.0 kg starts from rest and slides down a surface which corresponds to a quarter of a circle of 2.0 m radius.1)if the curved surface is smooth, find the speed at the bottom.2)if the speed at the bottom is 2.0m/s, calculate the energy dissipated due to friction in the descent.3)after the block reaches the horizontal with a speed of 2.0m/s it slides to a stop in a distance of 1.5m . Calculate the frictional force acting on the horizontal surface. Take g=10m/s2.

Answer 1

1) According to law of conservation of energy, in case without friction the terminal velocity at the end of the curve will be

$mgh=\frac{1}{2}mv^2,\\ \space\\ v=\sqrt{2gh}=\sqrt{2gR}=\sqrt{2\cdot9.8\cdot2}=6.26\text{ m/s}.$

2) Write law of conservation of energy:

$mgh=A_{\text{friction}}+\frac{1}{2}mu^2,\\ \space\\ A_{\text{friction}}=m\bigg(gR-\frac{1}{2}u^2\bigg)=70.4\text{ J}.\\$

3) The only force that affects block's motion is the force of friction. According to Newton's second law, the force of friction will be a net force:

$F_\text{net}=F_{\text{f}}=ma=m\cdot\frac{v^2}{2d}=4\cdot\frac{2^2}{2\cdot1.5}=5.33\text{ N}.$