The work done by a force acting on a body is as shown in the graph. The total work done in covering an initial distance of 20 m is

Question

The work done by a force acting on a body is as shown in the graph. The total work done in covering an initial distance of 20 m is <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/a4a82d3d59f0319871cab6a66b34255a-.png" /> (A) 225 J (B) 200 J (C) 400 J (D) 175 J

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/378bbb77655cf92164eb182ac92f585c-.png" /> Work done W= area under F-S graph = area of trapezium A B C D+ area of trapezium C E F D =(1/2) ×(10+15) × 10+(1/2) ×(10+20) × 5 =125+75=200 J

Q. The work done by a force acting on a body is as shown in the graph. The total work done in covering an initial distance of $20 m$ is

225 J

200 J

400 J

175 J

Work done $W =$ area under $F - S$ graph
$=$ area of trapezium $A BC D +$ area of trapezium $CEF D$
$= \frac{1}{2} \times (10 + 15) \times 10 + \frac{1}{2} \times (10 + 20) \times 5$
$= 125 + 75 = 200 J$

Q. The work done by a force acting on a body is as shown in the graph. The total work done in covering an initial distance of 20m is