Question

1. A box is along a frictionless horizontal floor and a force of 250 N pushes the box with an angle of 40^o with the horizontal. Starting from rest, the box achieves a velocity of 8 m/s in a time of 4 seconds. What is the weight of the box and the force exerted by the floor on the box?

Answer 1

Let's first find the acceleration of the box:

$$a=\dfrac{v-v_0}{t}=\dfrac{8\ \dfrac{m}{s}-0}{4\ s}=2\ \dfrac{m}{s^2}.$$

Applying the Newton's Second Law of Motion, we get:

$$F_{push}cos\theta=ma,$$ $$m=\dfrac{F_{push}cos\theta}{a}=\dfrac{250\ N\times cos40^{\circ}}{2\ \dfrac{m}{s^2}}=95.75\ kg.$$

Then, we can find the weight of the box:

$$W=mg=95.75\ kg\times 9.8\ \dfrac{m}{s^2}=938\ N.$$

Finally, we can find the force exerted by the floor on the box:

$$N=mg-F_{push}sin\theta=938\ N-250\ N\times sin40^{\circ}=777\ N.$$