A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7 N on a table with the coefficient of kinetic friction = 0.1. Compute the

(a) work done by the applied force in 10 s,

(b) work done by friction in 10 s,

(c) work done by the net force on the body in 10 s,

(d) change in kinetic energy of the body in 10 s,

Asked by Abhisek | 1 year ago |  115

##### Solution :-

The mass of the body = 2 kg

Horizontal force applied = 7 N

Coefficient of kinetic friction = 0.1

Acceleration produced by the applied force,

a1=$$\dfrac{F}{m}$$ =$$\dfrac{7}{2}$$ = 3.5 m/s2

Force of friction, f = μR = μmg = 0.1 x 2 x 9.8

Retardation produced by friction,

a2 = $$\dfrac{-f}{m}$$$$\dfrac{-196}{2}$$ = -0.98

Net acceleration  with which the body moves

a = a1 + a2 =  3.5 – 9.8 = 2.5

Distance moved by the body in 10 seconds,

s = ut + ($$\dfrac{1}{2}$$)at2 = 0 + ($$\dfrac{1}{2}$$) x 2.52 x (10)2= 126 m

(a) The time at which work has to be determined is t = 10 s

Work = Force x displacement

= 7 x 126 = 882 J

(b) Work done by the friction in 10 s

W = -f x s = -1.96 x 126

(c) Work done by the net force in 10 s

W = (F – f)s = (7 – 1.96) 126 = 635 J

(d) From v = u + at

v = 0 + 2.52 x 10 = 25.2 m/s

Final Kinetic Energy = ($$\dfrac{1}{2}$$) mv2

= ($$\dfrac{1}{2}$$) x 2 x (25.2)2 = 635 J

Initial Kinetic Energy= ($$\dfrac{1}{2}$$) mu2 = 0

Change in Kinetic energy = 635 – 0 = 635 J

The work done by the net force is equal to the final kinetic energy

Answered by Pragya Singh | 1 year ago

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