91Ó°ÊÓ

Speed of truck is given by,

$\begin{array}{c}v=130\text{ â¶Ä‰k³¾/³ó}\\ =36.1\text{ â¶Ä³¾/²õ}\end{array}$

Mass of truck,$M=1.2×{10}^4\text{kg}$

Inclination angle of road,

$\mathrm{θ}=15°$

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.

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

So, using thelaw of conservation of mechanical energy,

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

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

So,

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

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

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.

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

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