Answer to Question #197503 in Physics for Andu

Question

24. On a block of ice, located on a horizontal plane smooth at rest, acts a constant horizontal force of magnitude 40N for 5 sec. (f = 0). As a result of the action of force the block moves 100 m. What is the mass of the ice block?

Answer 1

The mass of the block can be found from the second Newton's law:

$$m = \dfrac{F}{a}$$

where $F = 40N$ is the force that acts on the block, and $a$ is its acceleration.

The distance covered by the block is given by the kinematic equation:

$$d = \dfrac{at^2}{2}$$

where $d = 100m$ is the distance, $t = 5s$ is the time. Expressing $a$ and substituting it into the first formula, obtain:

$$a = \dfrac{2d}{t^2}\\ m = \dfrac{Ft^2}{2d}\\ m = \dfrac{40\cdot 5^2}{2\cdot 100} = 5kg$$

Answer. 5kg.