Solveeit Logo

Question

Physics Question on work, energy and power

A point particle of mass mm, moves along the uniformly rough track PQRPQR as shown in the figure. The coefficient of friction, between the particle and the rough track equals μ\mu. The particle is released, from rest, from the point PP and it comes to rest at a point RR. The energies, lost by the ball, over the parts, PQPQ and QRQR, of the track, are equal to each other, and no energy is lost when particle changes direction from PQPQ to QRQR. The values of the coefficient of friction μ\mu and the distance x(=QR)x(=QR), are, respectively close to :

A

0.2 and 6.5 m

B

0.2 and 3.5 m

C

0.29 and 3.5 m

D

0.29 and 6.5 m

Answer

0.29 and 3.5 m

Explanation

Solution

From work energy theorem and given condition
mgh2μmgcosθhsinθ=0m g h-2 \,\mu m g \,\cos\, \theta \frac{h}{\sin \theta}=0
μ=12cot30=123=0.29\therefore \mu=\frac{1}{2 \cot 30}=\frac{1}{2 \sqrt{3}}=0.29
again mgh2=μmg.QR\frac{m g h}{2}=\mu \,m g \,. Q R
QR=h2μ=22×0.29=3.5m\therefore Q R=\frac{h}{2 \mu}=\frac{2}{2 \times 0.29}=3.5 \,m