Question

A ball of mass $m$ is thrown in air with speed $v_1$ from a height $h_1$ and it is at a height $h_2(>h_1)$ when its speed becomes $v_2$. Find the work done on the ball by the air resistance.

• A. $mg(h_2-h_1)+\frac{1}{2}m(v_2^2-v_1^2)$
• B. $mg(h_2+h_1)+\frac{1}{2}m(v_2^2-v_1^2)$
• C. $mg(h_2-h_1)$
• D. $\frac{1}{2}m(v_2^2-v_1^2)$

Answer 1

The correct option is A $mg(h_2-h_1)+\frac{1}{2}m(v_2^2-v_1^2)$

Work done on the ball by gravity is
$W_g=-mg(h_2-h_1)$
Let work done on the ball by air resistance be $W_{\text{air}}$
Applying Work Energy Theorem:
$W_g+W_{\text{air}}=\Delta KE$
$\Rightarrow -mg(h_2-h_1)+W_{\text{air}}=\frac{1}{2}m(v_2^2-v_1^2)$
$\Rightarrow W_{\text{air}}=mg(h_2-h_1)+\frac{1}{2}m(v_2^2-v_1^2)$