Question

In Figure 8-35, a runaway truck with failed brakes is moving downgrade at $\mathbf{130}\mathbf{}\mathit{k}\mathit{m}\mathbf{/}\mathit{h}$just before the driver steers the truck up a frictionless emergency escape ramp with an inclination of$\mathit{\theta }\mathbf{=}\mathbf{}\mathbf{15}\mathbf{°}$The truck’s mass is$\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}\mathbf{}}\mathit{k}\mathit{g}$

1. What minimum lengthmust the ramp have if the truck is to stop (momentarily) along it? (Assume the truck is a particle, and justify that assumption.) Does the minimum lengthincrease, decrease, or remain the same if
2. The truck’s mass is decreased and
3. Its speed is decreased?

Short answer

1. Minimum length L that the ramp must have if thetruck is to stop is $257m$.
2. If truck’s mass decreases, the L of the ramp does not decrease.
3. If truck’s speed decreases, L and h would both decrease.

Step-by-step solution

Step 1: Given

Speed of truck is given by,

$\begin{array}{c}v=130\text{\hspace{0.17em}\hspace{0.17em}km/h}\\ =36.1\text{\hspace{0.17em}\hspace{0.17em}m/s}\end{array}$

Mass of truck,$M=1.2\times {10}^4\text{kg}$

Inclination angle of road,

$\theta =15°$

Determining the concept

The problem is based on the law of conservation of energy which states that the total energy of an isolated system remains constant.According to the law of energy conservation, energy can neither be created, nor be destroyed.Using energy conversion, find the distance that the truck requires to stop.

Formula are as follow:

Kinetic energy is given by

$KE=\frac{1}{2}m{v}^2$

Potential energy is given by

$PE=mgh$

where, m is mass, v is velocity, g is an acceleration due to gravity and h is height.

(a) Determining the minimum length L that the ramp must have if the truck is to stop

At initial position, truck has only KE, and when it stops it has PE.

So, using thelaw of conservation of mechanical energy,

$\begin{array}{rcl}KE& =& PE\\ \frac{1}{2}m{v}^2& =& mgh\\ h& =& \frac{v^2}{2g}\\ h& =& \frac{{\left(36.1\right)}^2}{2\times 9.8}\\ h& =& 66.49m\\ & & \end{array}$

From theabove figure, the height that is found above is a vertical component of length L,

So,

$\begin{array}{rcl}h& =& L\sin \theta \\ L& =& \frac{h}{\sin \theta }\\ L& =& \frac{66.49}{\sin 15}\\ L& =& 256.78\\ L& =& 257m\end{array}$

Hence, minimum length L that the ramp must have if the truck is to stop is .

(b) Determining the length if the truck’s mass decreases

The above answer does depend on themass, so it will be thesame if mass is reduced.

Hence, if truck’s mass decreases, the L of the ramp does not decrease.

(c) Determining the length if the truck’s speed decreases

If the speed is decreased then L and h both would decrease.

Hence, if truck’s speed decreases, L and h would both decrease.