Question

The observer B finds that the work done by gravity is

• A. $\frac{1}{2}mg^{2}t_{0}^{2}$
• B. $-\frac{1}{2}mg^{2}t_{0}^{2}$
• C. $\frac{1}{2}mga_{0}t_{0}^{2}$
• D. $-\frac{1}{2}mga_{0}t_{0}^{2}$

Answer 1

Answer: (C)

$\frac{1}{2}mga_{0}t_{0}^{2}$

Answer 2

The gravitational force on the block is $F_g = mg$, acting vertically downwards. The displacement of the block is $s = \frac{1}{2}a_0t_0^2$, acting vertically downwards. Since the force of gravity and the displacement are in the same direction, the work done by gravity is positive.

$W_g = F_g \cdot s = mg \left(\frac{1}{2}a_0t_0^2\right) = \frac{1}{2}mga_0t_0^2$