The relationship between the force F and position a: 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 a: 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 (A) 30 J (B) 15 J (C) 25 J (D) 20 J

Tardigrade Admin · Accepted Answer

Work done = area enclosed by F-x graph. = area of OAM + area of ABNM + area of CDEN area of EFGH + area of HIJ =(1/2)× 1× 10+1× 10+1× 5-1× 5+(1/2)× 1× 10 =5+10+5-5+5=20J

Q. The relationship between the force F and position a: 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 enclosed by F-x graph.
= area of OAM + area of ABNM + area of CDEN area of EFGH + area of HIJ $= \frac{1}{2} \times 1 \times 10 + 1 \times 10 + 1 \times 5 - 1 \times 5 + \frac{1}{2} \times 1 \times 10$ $= 5 + 10 + 5 - 5 + 5 = 20 J$

Q. The relationship between the force F and position a: 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 enclosed by F-x graph.= area of OAM + area of ABNM + area of CDEN area of EFGH + area of HIJ =21​×1×10+1×10+1×5−1×5+21​×1×10 =5+10+5−5+5=20J

Work done = area enclosed by F-x graph.
= area of OAM + area of ABNM + area of CDEN area of EFGH + area of HIJ $= \frac{1}{2} \times 1 \times 10 + 1 \times 10 + 1 \times 5 - 1 \times 5 + \frac{1}{2} \times 1 \times 10$ $= 5 + 10 + 5 - 5 + 5 = 20 J$