Chapter 5.2, Problem 24PP

To determine

The time taken for the velocity of the box to become a given value.

Answer to Problem 24PP

The time taken for the velocity to double to 2 m/s is $t=0.5\text{ s}$.

Explanation of Solution

Given:

A force of 98-N is applied on a wooden box of mass 25-kg to move it along a wooden floor . The coefficient of kinetic friction between the two surfaces is 0.2.

Formula Used:

Here, a force is applied horizontally on the wooden box along the wooden floor . The entire distribution of forces on the box is shown in Figure 1.

  

Figure 1

There will be a force opposing the applied force which is due to kinetic friction. In this case, since the body is in motion, here, the body is in acceleration and hence there will be a net force in horizontal direction given by,

  $F_{net}=F_A-F_f$ …… (1)

The net force is given by

  $F_{net}=ma$

Now, the kinetic friction force will be proportional to the normal force acting in the body. In this case, the normal force is equal to the weight of the body as there are no other forces along the normal.

  $F_{\text{N}}=mg$ …… (2)

Thus, the kinetic friction force can be expressed as,

  $F_f={\mu }_{\text{k}}{F}_{\text{N}}$ …… (3)

Here, ${\mu }_{\text{k}}$ represents the coefficient of kinetic friction.

Based on this, the magnitude of horizontal force to be applied can be deduced as,

  $F_A=ma+{\mu }_{\text{k}}\left(mg\right)$ …… (4)

And, the acceleration on the body can be calculated as,

  $a=\frac{F_A-{\mu }_{\text{k}}\left(mg\right)}{m}$ …… (5)

Having found the acceleration, the time taken for the box to attain the given velocity from the initial value can be obtained as,

  $t=\frac{v_f-v_i}{a}$

Calculation:

Given the mass of bookcase as 25-kg, and horizontal force as 98 N.

Now, by substituting the acceleration due to gravity and given parameters, the acceleration on the box in the horizontal direction can be calculated based on (5) as

  $\begin{array}{c}a=\frac{98-0.2\left(25\times 9.8\right)}{25}\\ =1.96{\text{ m/s}}^2\end{array}$

Now, the time taken for the velocity of the box to attain 2 m/s from 1 m/s is obtained as,

  $\begin{array}{c}t=\frac{2-1}{1.96}\\ =0.5\text{ s}\end{array}$

Conclusion:

Thus, the time taken for the velocity to get doubled is 0.5 s.