Chapter 5.3, Problem 37PP

To determine

Magnitude of acceleration and final velocity of the crate.

Answer to Problem 37PP

Acceleration, $a=4.9{\text{ m/s}}^2$ final velocity, $v_f=19.6\text{ m/s}$ .

Explanation of Solution

Given:

A crate of 562 N weight is resting on a inclined at $\theta {\text{=30}}^{\text{o}}$ with the horizontal surface.

Initial velocity is 0 m/s since the crate is resting initially.

  $t_f=4\text{ sec}$

Formula used:

Force is given as,

  $F=ma$

Also, acceleration is equal to ratio of change in velocity to time.

Calculation:

Consider the figure shown below.

If the force is applied, the it will be equal to

  $\begin{array}{c}F=-F_{gx}\\ =-\left(-F_g\sin \theta \right)\\ =F_g\sin \theta \\ =mg\sin \theta \end{array}$

Rearrange the force formula for acceleration.

  $\begin{array}{l}a=\frac{F}{m}\\ a=\frac{mg\sin \theta }{m}\\ a=g\sin \theta \end{array}$

Substituting the given values,

  $\begin{array}{l}a=9.8\times \sin {30}^{\text{o}}\\ a=4.90{\text{ m/s}}^2\end{array}$

The final velocity will be

  $\begin{array}{l}a=\frac{v_f-v_i}{t_f-t_i}\\ 4.9=\frac{v_f-0}{4-0}\\ v_f=19.6\text{ m/s}\end{array}$

Conclusion:

Acceleration, $a=4.9{\text{ m/s}}^2$ final velocity, $v_f=19.6\text{ m/s}$ .