Chapter 4, Problem 58P

A crate is given an initial speed of 3.0 m/s up the 25.0° plane shown in Fig. 4-60. (a) How far up the plane will it go? (b) How much time elapses before it returns to its starting point? Assume ${\mu }_k$ = 0.12.

To determine

The distance the block would go up an inclined plane when given an initial velocity.

Answer to Problem 58P

Solution:

The distance the block goes up the block is 0.29 m.

Explanation of Solution

Given:

The initial speed of the block $v_0=3.0\text{ m/s}$

The angle of the incline $\theta =25.0°$

The speed of the block at the distance s is $v=0\text{ m/s}$

Formula used:

A block is given an initial speed. Itslides up a distance s along the length of the incline which is at an angle $\theta$ and comes to rest after traveling a distance sup to the plane.

The free body diagram for the block is shown in the diagram below.

  

The weight mg of the crate acts vertically downwards. The normal force $F_N$ acts perpendicular to the inclined plane. The weight mg is resolved into two components, $mg\sin \theta \text{ and }mg\cos \theta$ parallel and perpendicular to the inclined plane. The force of kinetic friction $F_{fr}$ acts opposite to the direction of motion of the box. There is a net force acting on the box in the downward direction parallel to the inclined plane accelerating it at a rate a .

Since there is no motion perpendicular to the incline,

  $F_N=mg\cos \theta$   ......(1)

The force of kinetic friction is related to the normal force as,

  $F_{fr}\text{=}{\mu }_k{F}_N$   ......(2)

Since the crate slides up the incline, there is a net force F that acts downwards. This is given by,

  $F=mg\sin \theta +F_{fr}$   ......(3)

Substitute equations (1) and (2) in (3).

  $\begin{array}{c}F=mg\sin \theta +{\mu }_{k}mg\cos \theta \\ =mg\left(\sin \theta +{\mu }_k\cos \theta \right)\end{array}$   .......(4)

From Newton’s second law,

  $F=ma$

Here, a is the resultant acceleration of the crate down the incline.

Therefore,

  $\begin{array}{c}ma=mg\left(\sin \theta +{\mu }_k\cos \theta \right)\\ a=g\left(\sin \theta +{\mu }_k\cos \theta \right)\end{array}$   ......(5)

The distance s it travels up the slope is calculated using the equation,

  $v^2=v_0^2+2as$

Calculation:

Calculate the acceleration of the crate by substituting 9.8 m/s²for g and the given values for ${\mu }_k$ and $\theta$ .

  $\begin{array}{c}a=g\left(\sin \theta +{\mu }_k\cos \theta \right)\\ =\left(9.8{\text{ m/s}}^2\right)\left[\left(\sin 25.0°\right)+\left(0.12\right)\left(\cos 25.0°\right)\right]\\ =5.207{\text{ m/s}}^2\end{array}$

Since the acceleration acts opposite to the direction of motion of the crate, a negative sign is affixed to the acceleration.

  $a=-5.207{\text{ m/s}}^2$

Calculate the distance the crate travels up the incline using the expression $v^2=v_0^2+2as$ and substituting the calculated value of a and the given values of v and $v_0$ in the equation,

  $\begin{array}{c}{v}^2=v_0^2+2as\\ {\left(0\text{ m/s}\right)}^2={\left(3.0\text{ m/s}\right)}^2+2\left(-5.207{\text{ m/s}}^2\right)s\\ s=\frac{\left(3.0\text{ m/s}\right)}{2\left(-5.207{\text{ m/s}}^2\right)}\\ =0.288\text{ m}\\ =0.29\text{ m}\end{array}$

Conclusion:

The motion of a block sliding up the inclined plane is analyzed by constructing its free body diagram. The acceleration of the block is determined using the free body diagram and Newton’s second law. Using the equations of motion, the distance the block slides up the plane is found to be 0.29m.

To determine

The time taken by the block t returns to the starting point.

Answer to Problem 58P

Solution:

The time taken by the block to return to the starting point is 1.2 s.

Explanation of Solution

Given:

The initial speed of the block $v_0=3.0\text{ m/s}$

The angle of the incline $\theta =25.0°$

The speed of the block at the distance s is $v=0\text{ m/s}$

Formula used:

The time taken by the crate to return to its starting point can be calculated using the third equation of motion.

  $s=v_{0}t+\frac{1}{2}a{t}^2$

Where, $v_o$ is the initial speed, t is the time, a is the acceleration and s is the displacement.

Calculation:

When the block reaches the starting point, its net displacement s_net is zero. Use 0 for s and the given values of v, $v_0$ , $\theta$ and $\left(9.8{\text{ m/s}}^2\right)$ for g in the equation $s=v_{0}t+\frac{1}{2}a{t}^2$ and solve for t .

  $\begin{array}{c}{s}_{net}=v_{0}t+\frac{1}{2}a{t}^2\\ \left(0\text{ m}\right)=\left(3.0\text{ m/s}\right)t+\frac{1}{2}\left(-5.207{\text{ m/s}}^2\right)t^2\end{array}$

The equation has two roots, $t=0$ and

  $\begin{array}{c}t=\frac{2\left(3.0\text{ m/s}\right)}{\left(5.207{\text{ m/s}}^2\right)}\\ =1.152\text{ s}\\ =1.2\text{ s}\end{array}$

Since $t=0$ refers to the time at which the block is set into motion, the time taken to reach the starting point is 1.2s.

Conclusion:

The motion of a block sliding upthe inclined plane is analyzed by constructing its free body diagram. The acceleration of the block is determined using the free body diagram and Newton’s second law. Using the equations of motion, the time taken by the block to reach the starting point is found to be1.2 s.