Question

A truck is traveling at 14.9 m/s down a hill when the brakes on all four wheels lock. The hill makes an angle of 11 ° with respect to the horizontal. The coefficient of kinetic friction between the tires and the road is 0.838. How far does the truck skid before coming to a stop?

Answer 1

A truck is traveling at $14.9\,\mathrm{m/s}$ down a hill when the brakes on all four wheels lock. The hill makes an angle of $11{}^{\circ}$ with respect to the horizontal. The coefficient of kinetic friction between the tires and the road is 0.838. How far does the truck skid before coming to a stop?

Solution:

$\mathrm{F_{fr}}$ - friction force;

$\alpha = 11{}^{\circ}$ - angle of the inclined plane;

$V_0 = 14.9\,\frac{\mathrm{m}}{\mathrm{s}}$ - initial velocity;

a - deceleration of the truck;

$\mu = 0.838$ - coefficient of kinetic friction;

S - distance that truck traveled;

Newton's second law for the truck on the X-axis:

x: $\mathrm{F_{fr} - mg_x = ma}$

Formula for the friction force:

$$\mathrm{F_{fr}} = \mu N \Rightarrow$$ $$\mu N - m g _ {x} = m a$$

Newton's second law for the truck on the Y-axis:

y: $N - m g_{y} = 0$

$$N = m g _ {y}$$

From the right triangle ABC:

$$\sin \alpha = \frac {m g _ {x}}{m g} \Rightarrow m g _ {x} = m g \sin \alpha$$ $$\cos \alpha = \frac {m g _ {y}}{m g} \Rightarrow m g _ {y} = m g \cos \alpha$$ $$N = m g _ {y} = m g \cos \alpha$$

(2)in(1):

$$\mu m g \cos \alpha - m g \sin \alpha = m a$$ $$\mu g \cos \alpha - g \sin \alpha = a$$ $$a = \mu g \cos \alpha - g \sin \alpha = 0. 8 3 8 \cdot 9. 8 \frac {\mathrm {m}}{\mathrm {s} ^ {2}} \cdot \cos 1 1 {}^ {\circ} - 9. 8 \frac {\mathrm {m}}{\mathrm {s} ^ {2}} \cdot \sin 1 1 {}^ {\circ} = 6. 2 \frac {\mathrm {m}}{\mathrm {s} ^ {2}}$$

Rate equation of the truck along X-axis:

$$z: 0 = V - a t$$ $$V = a t$$ $$t = \frac {V}{a}$$

Equation of motion for the truck along X-axis:

$$S = V t - \frac {a t ^ {2}}{2}$$

(4)in(5):

$$S = V \cdot \frac {V}{a} - \frac {a}{2} \cdot \left(\frac {V}{a}\right) ^ {2} = \frac {V ^ {2}}{a} - \frac {V ^ {2}}{2 a} = \frac {V ^ {2}}{2 a} = \frac {\left(1 4 . 9 \frac {\mathrm {m}}{\mathrm {s}}\right) ^ {2}}{2 \cdot 6 . 2 \frac {\mathrm {m}}{\mathrm {s} ^ {2}}} = 1 8 \mathrm {m}$$

Answer: distance that truck traveled, is equal to 18m.