Q. 24.1( 73 Votes )

# 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 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, and interpret your results.

Answer :

Given,

Mass of the body, m = 2kg

Force applied, F = 7N

Coefficient of kinetic friction, μ = 0.1

A. Time, t = 10s

Using Newton’s second law of motion,

F=ma

where, a is the acceleration caused by force F on a body of mass m

∴

⇒ a=3.5 m/s^{2}

Frictional force is given by

f=μmg

Where, μ is the coefficient of friction

m is the mass of the object

g is the acceleration due to gravity

∴ f=0.1 × 2kg × 9.8ms^{-2}

⇒ f=1.96N

Acceleration due to friction is given by

a_{f}=f/m

∴ (The negative sign is due to the fact that the frictional force acts opposite to the direction of motion)

⇒ a_{f}=-0.98ms^{-2}

Total acceleration = acceleration due to applied force + acceleration due to friction

Total acceleration, a=a_{F}+a_{f}

⇒ a=3.5ms^{-2}+(-0.98ms^{-2})

∴ a=2.52ms^{-2}

The displacement of the object is given by Newton’s second equation of motion as

⇒ s= 0 + (1/2)(2.52ms^{-2})(10s)^{2}

⇒ s=126 m

Work done by the applied force is given by

W=F∙s

⇒ W=7 N × 126 m

⇒ W=882 J

B. Work done by friction is given by

W=f∙s

⇒ W= (-1.96 N) × (126 m)

⇒ W= -246.96 J

C. Net force on the body, F_{t}=F+f

∴ F_{t}=7 N + (-1.96 N) = 5.04 N

Work done by total force is given by

W_{t}=F_{t}∙s

⇒ W_{t}=5.04 N × 126 m

⇒ W_{t}=635.04 J

D. Using newton’s First equation of motion, the final velocity is calculated as

v=u+at

⇒ v=0+(2.52 ms^{-2})(10 s)

⇒ v=25.2 m/s

Change in kinetic energy,

So, ΔK = ()(2kg)(25.2ms^{-1})^{2} – ()(2kg)(0)

⇒ ΔK = 635.04 J

This result is in accordance with Work-Energy theorem which states that the change in kinetic energy of an object is equal to the net work done on the object.

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