Question

A body of mass 2 kg initially at rest moves under the action of an applied force of 7 Newton on a table with coefficient of kinetic friction is 0.1 find work done frictional force in 10 seconds and work done by the net force in 10 seconds and also change in kinetic energy of the body ​

Answer 1

Answer:

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 ms-2

Frictional force is given as:

f = µmg

= 0.1 × 2 × 9.8 = – 1.96 N

The acceleration produced by the frictional force:

a” = -1.96 / 2 = -0.98 ms-2

Total acceleration of the body: a’ + a”

= 3.5 + (-0.98) = 2.52 ms-2

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

s = ut + (1/2)at2

= 0 + (1/2) × 2.52 × (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) mv2 – (1/2)mu2

= (1/2) × 2(v2 – u2) = (25.2)2 – 02 = 635 J

Answer 2

$\huge\mathfrak\red{Answer}$

ACCELERATION ACTS ON THE BODY

$mass = 2kg \\ force = 7 \: newton \\ \\ force = m \times acceleration \\ force =3 \times acceleration \\ \\ 7 = 3 \times acceleration \\ \\ accelertation = \frac{7}{3} m {sec}^{ - 2}$

DISPLACEMENT OF THE BODY IN 10SEC

$initial \: velocity(u) = 0 \\ acceleration = \frac{7}{3} m {sec}^{ - 2} \\ \\ displacement = \frac{1}{2} a {t}^{2} \\ \\ displacement = \frac{1}{2} \times \frac{7}{3} \times 100 \\ \\ displacement = \frac{350}{3} m$

WORK DONE BY FRICTIONAL FORCE

$force \: of \: friction = \gamma mg \\ where \: \gamma \: is \: the \: coefficient \: of \\ friction \\ \\ force \: of \: friction = 0.1 \times 2 \times 10 \\ \\ force \: of \: friction(f) = 2 \: newton \\ \\ work \: done \: by \: force = f.dr \\ \\ where \: dr \: is \: the \: displacement \: in \\ the \: direction \: of \: force \\ \\ work \: done = 2 \times \frac{350}{3} \\ \\ work \: done = - \frac{700}{3} joules \\ \\ where \: negative \: sign \: indicates \\ that \: force \: is \: in \: opposite \: direction$

NET FORCE ACTING ON THE BODY

$net \: force = 7 - 2 \\ \\ net \: force = 5newton \\ \\ work \: done \: by \: net \: force = f.dr \\ work \: done \: by \: net \: force = 5 \times \frac{350}{3} \\ \\ work \: done = \frac{1750}{3} joules$