A stockroom worker pushes a box with a mass 11.2 k g on a horizontal surface with a constant...

Question:

A stockroom worker pushes a box with a mass {eq}11.2 \ kg {/eq} on a horizontal surface with a constant speed of {eq}3.50 \ \frac{m}{s} {/eq}. The coefficient of kinetic friction between the box and the surface is {eq}0.20 {/eq}.

a) What horizontal force must be applied by the worker to maintain the motion?

b) If the force calculated in part (a) is removed, how far does the box slide before coming to rest?.

Newton's Second Law:

It states that the acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.

Answer and Explanation:

Part(a)

Since the box is moving at a constant velocity, the force exerted by the worker balances out with the friction such that there is no net force.

The force due to friction is \mu mg = 0.2(11.2)(9.81) = 21.9744N

Therefore the horizontal force is 21.9744N

Part(b)

If the force is removed, the deceleration is given by $$\frac{frictional \ \ force}{mass} = \frac{21.9744}{11.2} = 1.962m/s^2 $$

  • Initial velocity is 3.5m/s
  • Final velocity is 0m/s

$$\begin{align*} v^2 &= u^2 - 2as\\ s &= \frac{u^2 - v^2}{2a}\\ &= \frac{3.5^2 - 0}{2(1.962)}\\ &= 3.12m \end{align*} $$

Therefore the distance traveled is 3.12m


Learn more about this topic:

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Newton's Second Law: Meaning & Calculations

from Physics 111: Physics I

Chapter 6 / Lesson 7
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