Question

A skier moving at 5.47 m/sm/s encounters a long, rough, horizontal patch of snow having a coefficient of kinetic friction of 0.220 with her skis. How far does she travel on this patch before stopping?

Answer 1

Answer:

She travels 6.94 m before stopping.

Explanation:

Given:

Initial velocity of the skier (u) = 5.47 m/s

Final velocity of the skier (v) = 0 m/s (Comes to stop)

Coefficient of kinetic friction (μ) 0.220

Let the mass of skier be 'm',  acceleration of the skier be 'a' and the distance traveled be 'd' before coming to a stop.

Now, using the equation of motion as:

$v^2=u^2+2ad$

Plugging in the given values and expressing 'a' in terms of 'd', we get:

$0=(5.47)^2+2ad\\\\2ad=-29.92\\\\a=-\frac{29.92}{2d}=-\frac{14.96}{d}------(1)$

Now, the above acceleration is due to the frictional force. Now, frictional force is given as:

$f=\mu N\\\\f=\mu mg\\\\f=0.220\times m\times 9.8=2.156m$ -------------- (2)

Here, 'N' is the normal force which is equal to the weight of the skier.

From Newton's second law, frictional force is equal to the product of mass and acceleration. Here, friction acts in the opposite direction to motion.

So, $f=-ma=m(\frac{14.96}{d})$ ---------- (3)

Equating equations (2) and (3), we get:

$2.156m=\frac{14.96m}{d}\\\\d=\frac{14.96}{2.156}=6.94\ m$

Therefore, she travels 6.94 m before stopping.