Question

An automobile moving at a speed of $72km\,h$ reaches the foot of a smooth incline when the $y$ engine is switched off. How much distance does the automobile go up the incline before stopping? the incline makes an angle of $30^\circ $ with the horizontal. Take $8 = 9.8ms$.

Answer 1

Hint: If a particular object is falling, the particle is bound to point in the direction of gravity. The magnitude of the falling body depends on the mass, gravitational constant and height from which it is falling. The force applied to the object is an external force, from outside the system. When it does positive work It increase the gravitational potential energy of the system.

Complete step by step answer:

Here, \[40.8m\]$u = 72km/h = \dfrac{{72 \times 1000}}{{60 \times 60}}m/s$
$u = 20m/s$
$\theta = 30^\circ $
$g = 9.8m/{s^2}$
$S = ?$
$v = 0$
Acceleration $a = - g\sin \theta $
$
  = - 9.8\sin 30^\circ \\
  \therefore a = - 4.9m/{s^2} \\
 $
$ \Rightarrow {v^2} - {u^2} = 2as$
$v$ is final velocity, $u$ is initial velocity and $a$is acceleration $S$is distance.
$
  0 - 20 \times 20 = 2( - 4.9) \\
  S = \dfrac{{ - 400}}{{ - 9.8}} \\
   = 40.8m \\
$
The distance traveled by automobile on incline before going to rest is \[ = 40.8m\]

Additional Information:
The gravitational potential at a point in a gravitational field is the work done per unit mass that would have to be done by some externally applied force to bring a massive object to that point from some defined position of zero potential, is due to its position relative to the surroundings within the earth object system.

Note:
Work done by a body is against the applied force if the displacement is in a direction opposite to the force. Work done against gravity $ = $Force$ \times $Displacement$ = $Weight$ \times $height $ = $mgh$ = $gravitational potential energy, stored in the body. When the force is constant, the work done is defined as the product of the force and distance moved on the direction of force.