Question

A body of 5 kg mass placed on a smooth
horizontal surface experiences a force varying
with displacement of body as shown in figure.
The Kinetic energy of body when force ceases
to act on it is​

Answer 1

Answer:

4000J

Explanation:

Given, mass of the body, m=5kg;

Initial velocity of the body, u=0;

Force acting on the body, F=20N;

Time for which the force acts, t=10s

If a is the acceleration produced in the body,

a=

m

F

=

5

20

=4m/s

2

Let v be the velocity of the body after 10 s.

Clearly, v=u+at=0+(4)×(10)=40m/s

Kinetic energy acquired by the body,

E

k

=

2

1

mv

2

=

2

1

(5)×(40)

2

=4000J

Answer 2

Given:

A body of 5 kg mass placed on a smooth horizontal surface experiences a force varying with displacement of body as shown in figure.

To find:

The Kinetic energy of body when force ceases

to act on it is ?

Calculation:

In these types of questions, its best to apply WORK-ENERGY THEOREM, where the total work done by the force is equal to the change in kinetic energy.

• The work can be calculated by the area under F vs x graph.

Applying the theorem:

$\sf \therefore \: work = \Delta KE$

$\sf \implies \: area \: of \: graph = \Delta KE$

$\sf \implies \: (\frac{1}{2} \times 10 \times 20) + (5 \times 10) + ( \frac{1}{2} \times 5 \times 10) + (5 \times 10) + ( \frac{1}{2} \times 5 \times 10) = \Delta KE$

$\sf \implies \: 100 + 50 + 25 + 50 + 25= \Delta KE$

$\sf \implies \: \Delta KE = 250$

$\sf \implies \: KE_{f} - KE_{i}= 250$

$\sf \implies \: KE_{f} - 0= 250$

$\sf \implies \: KE_{f} = 250 \: joule$

So, final KINETIC ENERGY is 250 J.