Question

A string is attached to a body of mass 2 kg which is at rest on the ground. If the string  is pulled vertically upward with a constant force of 42 N, find how far the body rises  during the first second.

Answer 1

The total force on the body is given as follows:

$$F = 42N-mg$$

where $mg$ is the weight of the body ($m = 2kg, g= 9.8m/s^2$ ) and "-" sign is because the weight is directed opposite to the pulling force. Thus, obtain:

$$F = 42N - 2kg\cdot 9.8 = 22.4N$$

According to the second Newton's law, the acceleration of the body is:

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

And according to the kinematic law, the distance travelled during $t=1s$ is:

$$d = \dfrac{at^2}{2} = \dfrac{Ft^2}{2m}\\ d = \dfrac{22.4N\cdot (1s)^2}{2\cdot 2kg} = 5.6m$$

Answer. 5.6 m.