A force of `10N` is acting on a block of `20kg` on horizontal surface with coefficient of friction `mu=0.2`. Calculate the work done by the force.

A block of mass 2 kg initially at rest moves under the action of an applied horizontal force of 6 N on a rough horizontal surface. The coefficient of friction between block and surface is 0.1. The work done by the applied force in 10 s is (Take g = 10 ms^(-2) )

A 100 N force acts horizontally on a block of 10 kg placed on a horizontal rough surface of coefficient of friction mu=0.5 . If the acceleration due to gravity (g) is taken as 10ms^(-2) , the acceleration of the block (in ms^(-2) ) 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 horizontal force of 129.4 N is applied on a 10 kg block which rests on a horizontal surface. If the coefficient of friction is 0.3, the acceleration should be

A block of mass 10kg is placed on a rough horizontal surface having coefficient of friction mu=0.5 . If a horizontal force of 100N is acting on it, then acceleration of the will be.

A force of 100N is applied on a block of mass 3kg as shown in the figure. The coefficient of friction between the surface and the block is mu=(1)/(sqrt(3) . The friction force acting on the bock is.

A force F is acting at an angle theta with horizontal on a block of mass 12 kg placed on a rough 3 horizontal surface having coefficient of friction mu=3/ 4 as shown in the figure. The minimum value of force F to just move the block will be (g = 10 ms^-2)

A block of mass m is kept over another block of mass 2m and the system rests on a smooth horizontal surface. The coefficient of friction between the block is 0. 50 . Find the work done by the force of friction on the smaller block by the bigger block during a displacement the of the system when a force mg is applied to the lower block

A rectangular block of mass 5 kg is kept on a horizontal surface . The coefficient of friction between the block and the surface is 0.2 . If a force of 20 N is applied to the block at angle of 30^(@) with the horizontal plane , what is the force of friction on the block ? (Take g = 10 m//s^(2) )

A block of mass m=5kg is resting on a rough horizontal surface for which the coefficient of friction is 0.2 . When a force F=40N is applied, the acceleration of the block will be (g=10m//s^(2)) .