A 14 kg box slides down a long, frictionless incline of angle 30 degree . It starts from rest at...

Question:

A {eq}14 \ kg {/eq} box slides down a long, frictionless incline of angle {eq}30^\circ {/eq}. It starts from rest at time {eq}t = 0 {/eq} at the top of the incline at a height of {eq}22 \ m {/eq} above ground.

(a) What is the original potential energy of the box relative to the ground?

(b) From Newton's laws, find the distance the box travels in {eq}1 \ s {/eq} and its speed at {eq}t = 1 \ s {/eq}.

(c) Find the potential energy and the kinetic energy of the box at {eq}t = 1 \ s {/eq}.

(d) Find the kinetic energy and the speed of the box just as it reaches the bottom of the incline.

Newton's Laws of Motion:

From the information given, it is clear that the block slides down due to gravity but because there is an incline, you use trigonometry to find the component of the acceleration down the slope. If you are confused about which equation from Newton's Laws to apply, simply list down the details you have and compare with the equations.

Answer and Explanation:

Part(a)

The potential energy {eq}E_{p} {/eq} is computed as follows

$$\begin{align*} E_{p} &= mgh\\ &= 14(9.81)(22) = 3021.5J \end{align*} $$

Part(b)

Considering that down the incline, {eq}a = g\sin \theta {/eq} and t = 1s,

$$\begin{align*} v &= u + at\\ &= 0 + gt\sin \theta\\ &= 9.81(1)\sin 30 = 4.9m/s \end{align*} $$

the distance at 1s is calculated as follows

$$\begin{align*} v^2&= u^2 +2as \\ &= 0 + 2s_{1}g\sin \theta\\ s_{1}&= \frac{v^2}{2g \sin \theta}\\ &= \frac{4.9^2}{2(9.81) 30}\\ &= 2.45m \end{align*} $$

Part(c)

The new perpendicular height {eq}h_{1} = h - s_{1} \sin\theta {/eq}. The potential energy {eq}E_{p1} {/eq} is computed as follows

$$\begin{align*} E_{p1} &= mgh_{1}\\ &= 14(9.81)(22 - 2.55\sin 30) = 2846.4J \end{align*} $$

The kinetic Energy is calculated as follows

$$\begin{align*} E_{k1} &= \frac{1}{2}mv_{1}^2\\ &= \frac{1}{2}(14)4.9^2 = 168.1J \end{align*} $$

Part (d)

By the law of conservation of energy, the final kinetic energy just before the box reaches the bottom is equal to the potential energy we started with.

$$\therefore E_{kf} = 3021.5J $$

Now,

$$\begin{align*} E_{kf} &= \frac{1}{2}mv_{f}^2\\ v_{f}&= \sqrt\frac{2E_{kf}}{m}\\ &= \sqrt\frac{2(3021.5)}{14}\\ &=20.8m/s^2 \end{align*} $$


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