Chapter 6, Problem 73GP

To determine

The final speed of the crate

Answer to Problem 73GP

Solution: $\text{14}\text{.9 }\frac{\text{m}}{\text{s}}$

Explanation of Solution

Given The mass of the crate is $m=36.0\text{kg}$

The initial speed of the crate is $v_0=0$

The constant horizontal force is $F=225\text{N}$

The distance moved by the crate on the frictionless floor is 11.0 m.

Formula used

Third equation of motion:

  $v^2=v_0{}^2+2as$

Here, v is the final velocity, u is the initial velocity, a is the acceleration, s is the distance moved.

For the surface with friction, the acceleration is $a_1=(F-\mu mg)/m$ , here $\mu mg$ is the force of friction. This is because the force of friction acts in a direction opposite to the applied force ( F ).

Calculations

The acceleration of the crate on the frictionless surface is:

  $\begin{array}{l}a=\frac{F}{m}\\ a=\frac{225}{36.0}\frac{\text{m}}{{\text{s}}^{\text{2}}}\\ a=6.25\frac{\text{m}}{{\text{s}}^{\text{2}}}\end{array}$

Since the crate starts from rest, hence the velocity after moving 11.0 m is:

  $\begin{array}{l}{v}^2=0+2\times 6.25\times 11.0\\ v=11.7\frac{\text{m}}{\text{s}}\end{array}$

The acceleration of the crate on the surface with friction is,

  $\begin{array}{l}{a}_1=(F-\mu mg)/m\\ a_1=\frac{225-0.20\times 36\times 9.8}{36.0}\frac{\text{m}}{{\text{s}}^{\text{2}}}\\ a_1=4.30\frac{\text{m}}{{\text{s}}^{\text{2}}}\end{array}$

Therefore, the final velocity of the crate for the next 10.0 m is calculated from the kinematics equation $v_f{}^2=v_i{}^2+2a{s}_1$

  $\begin{array}{l}{v}_1{}^2=v^2+2a_1\times 10.0\\ $ v_1{}^2={(11.7)}^2+2\times 4.30\times 10.0\frac{{\text{m}}^{\text{2}}}{{\text{s}}^{\text{2}}}$v_1{}^2=223\frac{{\text{m}}^{\text{2}}}{{\text{s}}^{\text{2}}}\\ v_1=14.9\frac{\text{m}}{\text{s}}\end{array}$

Conclusion:

The final speed of the crate can be calculated from the consideration of the equations of motion.