Question

It takes 5 seconds for a 2 kg box to be pushed 10 meters from rest. What was the forceof the push?

Answer 1

Given data:

* The mass of the box is 2 kg.

* The time taken by the box to travel the given distance is 5 seconds.

* The distance traveled by the box is 10 meters.

* The initial velocity of the box is 0 m/s.

Solution:

By the kinematics equation, the distance traveled by the box in terms of its acceleration is,

`S=ut+(1)/(2)at^2`

where u is the initial velocity, t is the time taken, a is the acceleration, and S is the distance traveled,

Substituting the known values,

`\begin{gathered} 10=0+(1)/(2)* a*(5)^2 \\ 10=(25)/(2)* a \\ a=10*(2)/(25) \\ a=0.8ms^(-2) \end{gathered}`

By the Newton's second law, the force exerted on the box in terms of the acceleration is,

`F=ma`

where m is the mass of the box, a is the acceleration and F is the force,

Substituting the known values,

`\begin{gathered} F=2*0.8 \\ F=1.6\text{ N} \end{gathered}`

Thus, the force of the push is 1.6 N.