Question

A small block with mass 0.0475 kg slides in a vertical circle of radius 0.425 m on the inside of a circular track. During one of the revolutions of the block, when the block is at the bottom of its path, point A, the magnitude of the normal force exerted on the block by the track has magnitude 3.95 N . In this same revolution, when the block reaches the top of its path, point B, the magnitude of the normal force exerted on the block has magnitude 0.670 N. How much work was done on the block by friction during the motion of the block from point A to point B?

Answer 1

Answer:0.10283 J

Explanation:

Given

mass of block(m)=0.0475 kg

radius of track (r)=0.425 m

when the Block is at Bottom Normal has a magnitude of 3.95 N

Force acting on block at bottom

$N-mg=\frac{mu^2}{r}$

$N=mg+\frac{mu^2}{r}$

$3.95=0.0475\timess 9.81+\frac{0.0475\times u^2}{0.425}$

$u^2=31.172$

$u=\sqrt{31.172}$

u=5.583 m/s

At top point

$N+mg=\frac{mv^2}{r}$

$0.670+0.0475\timess 9.81=\frac{mv^2}{r}$

$v^2=10.1639$

v=3.188 m/s

Using Energy conservation

$\frac{mu^2}{2}=\frac{mv^2}{2}+mg\left ( 2r\right )+W_f$

$W_f=0.4989-0.39607$

$W_f=0.10283 J$

i.e. 0.10283  J of energy is wasted while moving up.