A scooter weighs 120 kg f. Brakes are applied so that wheels stop rolling and start skidding. Find the force of friction if the coefficient of friction is `0.4`.

In Fig. 6-48, a 2.0 kg block lies on a 20 kg trolley that can roll across a floor on frictionless bearings. Between the block and the trolley, the coefficient of kinetic friction is 0.20 and the coefficient of static friction is 0.25. When a horizontal 2.0 N force is applied to the block, what are the magnitudes of (a) the frictional force between the block and the trolley and (b) the acceleration of the trolley?

A force F is applied at the centre of a disc of mass M. The minimum value of coefficient of friction of the surface for rolling is

A force of 49 N just able to move a block of wood weighing 10 kg on a rough horizontal surface , then coefficient of friction is

A 30kg block rests on a rough horizontal surface A force of 200N is applied on the block The block acquires a speed of 4m//s starting from rest in 2s What is the value of coefficient of friction ? .

Assertion: When brakes are applied on a wet road, a car is likely to skid. Reason: Brakes prevent rotation of the wheels, and there is not sufficient friction between the road and the whells.

A solid flywheel of 20 kg mass and 120 mm radius revolves at 600 "revmin"^(-1) . With what force must a brake lining be pressed against it for the flywheel to stop in 3s, if the coefficient of friction is 0.1 ?

A block of mass m = 2kg is accelerating by a force F = 20N applied on a smooth light pulley as shown in the figure. If the coefficient of kinetic friction between the block and the surface is mu=0.3 , find its acceleration.

A force of 98 N is required to just start moving a body of mass 100 kg over ice. The coefficient of static friction is