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# Solution - 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 - CBSE (Science) Class 11 - Physics

ConceptIntroduction of Work, Energy and Power

#### Question

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.

#### Solution 1

Mass of the body, m = 2 kg

Applied force, F = 7 N

Coefficient of kinetic friction, µ = 0.1

Initial velocity, u = 0

Time, t = 10 s

The acceleration produced in the body by the applied force is given by Newton’s second law of motion as:

a' =F/m = 7/2 = 3.5 "m/s"^2

Frictional force is given as

f = mumg

= 0.1 xx 2xx 9.8 = - 1.96 N

The acceleration producd by the frictional force

a"= 1.96/2 = -0.98 "m/s"^2

Total acceleration of the body:
a = a' + a"

=3.5 + (-0.098) = 2.52 "m/s"^2

The distance travelled by the body is given by the equation of motion:

s = ut + 1/2 at^2

= 0 + 1/2 xx 2.52 xx (10)^2

= 126 m

a) Work done by the applied force, Wa = F × s = 7 × 126 = 882 J

(b) Work done by the frictional force, Wf = F × s = –1.96 × 126 = –247 J

(c) Net force = 7 + (–1.96) = 5.04 N

Work done by the net force, Wnet= 5.04 ×126 = 635 J

(d) From the first equation of motion, final velocity can be calculated as:

v = u + at

= 0 + 2.52 × 10 = 25.2 m/s

Change in kinetic energy = 1/2 mv^2 - 1/2 mu^2

= 1/2xx2(v^2-u^2) =  (25.2)^2- 0^2 = 635 J

#### Solution 2

(a) We know that Uk = frictional force/normal reaction frictional force = Uk x normal reaction

= 0.1 x 2 kg wt = 0.1 x 2 x 9.8 N = 1.96 N

net effective force = (7 – 1.96) N = 5.04 N

acceleration = 5.04/2 ms-2 = 2.52 ms-2

distance, s=1/2x 2.52 x 10 x 10 = 126 m

work done by applied force = 7 x 126 J = 882 J

(b) Work done by friction = 1.96 x 126 = -246.96 J

(c) Work done by net force = 5.04 x 126 = 635.04 J

(d) Change in the kinetic energy of the body = work done by the net force in 10 seconds = 635.04 J (This is in accordance with work-energy theorem).

Is there an error in this question or solution?

#### APPEARS IN

NCERT Physics Textbook for Class 11 Part 1 (with solutions)
Chapter 6: Work, Energy and Power
Q: 2 | Page no. 134

#### Reference Material

Solution for question: 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 concept: Introduction of Work, Energy and Power. For the courses CBSE (Science), CBSE (Arts), CBSE (Commerce)
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