Question
A force $${F_x}$$ acts on a particle such that its position $$x$$ changes as shown in the figure. The work done by the particle as it moves from $$x = 0$$ to $$20\,m$$ is
A force $${F_x}$$ acts on a particle such that its position $$x$$ changes as shown in the figure. The work done by the particle as it moves from $$x = 0$$ to $$20\,m$$ is
A.
$$37.5\,J$$
B.
$$10\,J$$
C.
$$45\,J$$
D.
$$22.5\,J$$
Answer :
$$45\,J$$
Solution :
$$\eqalign{ & W = {\text{area of }}F - x{\text{ graph}} \cr & = {\text{area of }}\Delta + {\text{area of rectangle}} + {\text{area of }}\Delta \cr & = \frac{{5 \times 3}}{2} + 10 \times 3 + \frac{{5 \times 3}}{2} = 45\,J \cr} $$
$$\eqalign{ & W = {\text{area of }}F - x{\text{ graph}} \cr & = {\text{area of }}\Delta + {\text{area of rectangle}} + {\text{area of }}\Delta \cr & = \frac{{5 \times 3}}{2} + 10 \times 3 + \frac{{5 \times 3}}{2} = 45\,J \cr} $$