Question

A ball is released from top of tower. The ratio of work done by force of gravity in first,second and third second of motion of ball is
(1)1:2:3
(2)1:4:9
(3)1:3:5
(4)1:5:3

Answer 1

Answer on Question #49237 – Physics – Mechanics | Kinematics | Dynamics

1. A ball is released from top of tower. The ratio of work done by force of gravity in first, second and third second of motion of ball is

(1) 1:2:3

(2) 1:4:9

(3) 1:3:5

(4) 1:5:3

Solution.

The motion of a ball is with the constant acceleration $g$ . So, the height depends on the time as $h = \frac{g t^2}{2}$ .

Let denote $t_0 = 1$ s.

The work done the gravity force in the first second: $A_{1} = F\cdot \Delta h = mg\cdot \frac{g t_{0}^{2}}{2} = \frac{mg^{2}t_{0}^{2}}{2}$

The work done the gravity force in the first second: $A_{2} = mg\cdot \left[\frac{g(2t_{0})^{2}}{2} -\frac{g t_{0}^{2}}{2}\right] = \frac{3mg^{2}t_{0}^{2}}{2}$

The work done the gravity force in the first second: $A_{3} = mg\cdot \left[\frac{g(3t_{0})^{2}}{2} -\frac{g(2t_{0})^{2}}{2}\right] = \frac{5mg^{2}t_{0}^{2}}{2}$

One can see that $A_{1}:A_{2}:A_{3} = 1:3:5$

Answer: 3)

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