The relationship between the force F and position x of a body is as shown in figure. The work done in displacing the body from x = 1 m to x = 5 m will be

Question

The relationship between the force F and position x of a body is as shown in figure. The work done in displacing the body from x = 1 m to x = 5 m will be <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/58da8ff508aeb476bb02f09159fd2bd8-.png" /> (A) 30 J (B) 15 J (C) 25 J (D) 20 J

Tardigrade Admin · Accepted Answer

Work done = area under F-x graph = area of rectangle ABCD + area of rectangle LCEF + area of rectangle GFIH + area of triangle IJK <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/88be3104088fd0a017cb4363afea4fc9-.png" /> =(2-1)×(10-0)+(3-2)(5-0)+(4-2)(-5-0) +(1/2)(5-4)(10-0)=15 J

Q. The relationship between the force $F$ and position $x$ of a body is as shown in figure. The work done in displacing the body from $x = 1 m$ to $x = 5 m$ will be

30 J

15 J

25 J

20 J

Work done = area under $F - x$ graph

= area of rectangle $A BC D +$ area of rectangle $L CEF$

+ area of rectangle $GF I H +$ area of triangle $I J K$

$= (2 - 1) \times (10 - 0) + (3 - 2) (5 - 0) + (4 - 2) (- 5 - 0)$

$+ \frac{1}{2} (5 - 4) (10 - 0) = 15 J$

Q. The relationship between the force F and position x of a body is as shown in figure. The work done in displacing the body from x=1m to x=5m will be