Question

A body of mass 2 kg is resting on a rough horizontal surface a force of 20 newton is applied to wait for 10 seconds parallel to the surface if the coefficient of kinetic friction between the surfaces in contact is 0.2 calculate first part work done by the applied force in 10 seconds be change in kinetic energy of the object in 10 seconds equal to 10 metre per second squared

Answer 1

m = 2 kg                 F = 20 N                   t = 10 s

Friction force = μ m g = 0.2 * 2 * 10 = 4 N

Net force in forward direction = 20 - 4 = 16 N
So acceleration = a = Force / m  = 16 /2 = 8 m/s²

Initial speed = u = 0
Distance traveled = s = u t + 1/2 a t² = 0 + 1/2 * 8 * 10² = 400 m

Work done by F = F * s = 20 * 400 = 8 kJ
Work done by Friction force = - 4 * 400 = - 1.6 kJ  (heat dissipated)

Kinetic energy of the body = 1/2 m v² = 1/2 m (2 a s)
      = 2 * 8 * 400 = 6.4 kJ

Answer 2

frictional force= 0.2 ×2×10=4N ; net force= 20-4=26N a=16/2=8m/s^2. s= ut+1/2at^2= 1/2×8×100= 409m. W=fs=20×400=8000j. no w, v^-u^2=2as= v^2=2as. KE = 1)2mv^2= 1/2×2×2×8×400=6400j