Question

A truck is traveling at 14.6 m/s down a hill when the brakes on all four wheels lock. The hill makes an angle of 18 ° with respect to the horizontal. The coefficient of kinetic friction between the tires and the road is 0.915. How far does the truck skid before coming to a stop?

Answer 1

Answer on Question#37445 – Physics – Mechanics

We will use kinematic equation

$$v _ {f} ^ {2} = v _ {0} ^ {2} + 2 a d,$$

where $v_{f} = 0$ - final velocity, $v_{0} = 14.6\frac{\mathrm{m}}{\mathrm{s}}$ - initial velocity, $a$ - acceleration, $d$ - distance.

The forces acting on the truck once the brakes are applied are the component of gravity down the hill, $mg \sin \theta$, and the friction acting up the hill, $\mu mg \cos \theta$ where $\mu$ is the coefficient of kinetic friction. Therefore, choosing down the hill as the positive direction, we have

$$m a = m g \sin \theta - \mu m g \cos \theta ,$$

or

$$a = g \sin \theta - \mu g \cos \theta .$$

So

$$d = - \frac {v _ {0} ^ {2}}{2 a} = \frac {v _ {0} ^ {2}}{\mu g \cos \theta - g \sin \theta},$$ $$d = \frac {1 4 . 6 ^ {2}}{9 . 8 1 * (0 . 9 1 5 * \cos 1 8 {}^ {\circ} - \sin 1 8 {}^ {\circ})} = 3 8. 7 m.$$

Answer: 38.7 m.