CameraIcon

SearchIcon

MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Work Energy Theorem Application

A body moves ...

Question

A body moves from point A and B under the action of a force, varying in magnitude as shown in figure below. Obtain the work done. Force is expressed in newton and displacement is in metre.

Open in App

Solution

Work done=Area enclosed from A to B
=Area enclosed from A to L(Since area from L to M and M to B are equal and opposite polarity)
$= 10 \times 2 + \frac{1}{2} \times 1.5$
$= 20 + 2.5 = 22.5 J$

Suggest Corrections

thumbs-up

1

Similar questions

Q. A particle moves from a point

_{1} = 3^i + 2^j - 6^k

_{2} = 14^i + 13^j - 9^k

under the action of a force

\to F = 4^i +^j + 3^k

newton. If the displacement between two points is in metres, calculate the work done.

Q. A horizontal force of 20 newton is required to keep a body moving in horizontal direction with constant velocity. What is the work done by the frictional force and the displacement of the body is 2 metre.

Q. The force vs displacement graph of a moving body is as shown in the figure. What is the work done as the body moves from

x = 5 m

x = 35 m

Q. Under the action of a force, a

2 k g

body moves such that its position

x

as a function of time is given by

x = \frac{t^{3}}{3}

x

is in metre and

t

in second. The work done by the force in the first

2 s

Q. Under the action of a force, a

2 l g

body moves such that it position

x

as a function of time is given by

x = \frac{t^{3}}{3}

x

is in metre and

t

in seconds. The work done by the force in the first two seconds is

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Conservative Forces and Potential Energy

PHYSICS

Watch in App

explore_more_icon

Explore more

Work Energy Theorem Application

Standard XII Physics

Join BYJU'S Learning Program

CrossIcon