Physics, asked by Sajid1347, 10 months ago

a block of mass 1 kgmoves on a horizontalrough surface under the application of 10 N horizontal force.If the magnitude of frictional force acting on block is 2 N. Calculate the work done by the gravity,10 N force, frictional force,normal reaction force at the end of 2 seconds​

Answers

Answered by MukulCIL
0

Explanation:

Work is Force multiplied by the displacement in the direction of the force.

Since the surface is horizontal and so is the displacement, work done by gravitational force anf normal force will be zero.

Work done by 10 N force will be

= Force X Displacement Cos of angle between force and displacement

Drawing FBD of the body we will get Force of 10N and friction of 2N in the horizontal plane .

Net horizontal force will be 8N and mass is 1 kg then acceleration is direction of displacement will be = 8/1 = 8 m/s^2.

S = ut + 1/2 at^2------------( using this because acceleration is constant)

u= 0

S = .5 * 8 * 4

S = 16 m----- diaplacement

Work by 10N force

= 10*16 = 160 JOULE

Work by frictional force

= -2 * 16 = -32 Joule

negativr indicates both force and displacement are opposite to each other.

Answered by geniusno0001
0

Answer:

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