Question

A wheel of radius 20 cm is pushed to move it on a rough horizontal surface. It is found to move through a distance of 60 cm on the road during the time it completes one revolution about the center. Assume that the linear and the angular accelerations are uniform. The frictional force acting on the wheel by the surface is

(a) along the velocity of the wheel
(b) opposite to the velocity of the wheel
(c) perpendicular to the velocity of the wheel
(d) zero

Answer 1

A ALONG TO VELOCITY BECAUSE WHEN IT ROTATED THE VELOCITY INCREASE SO ANSWER IS A

Answer 2

Answer ⇒ Option (a) . Along the Velocity of the Wheel.

Explanation ⇒

Radius of the wheel(r) = 0.20 m.

We know that the Perimeter of the wheel will be given by the Relation,

P = 2πr

= 2π × 0.20

= 0.4 × 22/7

= 1.26

But According to the Question, it has covered the distance of only 0.6 m.

Therefore, the wheel must have slips on the road. At the point of contact with the road, the part of the wheel is sliding backward.

Hence the force of friction acts forward i.e. along the velocity of the wheel.

Hence, Option (a). is correct.

Hope it helps.