Question

A constant force $10\text{N}$ acts on a ball of mass $2\text{kg}$ for $2\text{s}$ . Then another force $18\text{N}$ acts on the same ball for $\frac{10}{9}\text{second}$. If ball is initially at rest, then find the total work done.

• A. $400\text{J}$
• B. $300\text{J}$
• C. $200\text{J}$
• D. $100\text{J}$

Answer 1

The correct option is A $400\text{J}$

When $F_1=10\text{N}$ acts on the ball
$a_1=F_1/m=5{\text{m/s}}^2$
$v_1=u_o+at$ $=0+5\times 2$ $(\because u_o=0)$
$v_1=10\text{m/s}$
Now,
$v_1^2-u_o^2=2a{S}_1$
$\Rightarrow {10}^2-0^2=2\times 5\times S_1$
$\Rightarrow S_1=10\text{m}$
Work done by $F_1=F_1\times S_1=10\times 10$
$W_1=100\text{J}$

Again,
when $F_2=18\text{N}$ acts on ball,
$a_2=F_2/m=\frac{18}{2}=9{\text{m/s}}^2$

$S_2=v_{1}t+\frac{1}{2}a{t}^2=10\left(\frac{10}{9}\right)+\frac{1}{2}\left(9\right)(\frac{10}{9}{)}^2=\frac{150}{9}=\frac{50}{3}$
Work done by force $W_2=F_2\times S_2$
$W_2=\frac{50}{3}\times 18=300\text{J}$

Total work done = $W_1+W_2=300+100=400\text{J}$