Question

Under the influence of its drive force, a snowmobile is moving at a constant velocity along a horizontal patch of snow. When the drive force is shut off, the snowmobile coasts to a halt. The snowmobile and its rider have a mass of 134 kg. Under the influence of a drive force of 205 N, it is moving at a constant velocity whose magnitude is 6.36 m/s. The drive force is then shut off. Find (a) the distance in which the snowmobile coasts to a halt and (b) the time required to do so.

Answer 1

(a) 13.2 m

When the drive force is on, the snowmobile is moving at constant velocity: this means that the acceleration is zero, so the net force is zero. But the net force is the resultant of two forces acting on the snowmobile: the driving force, F, and the frictional force, Ff:

$F-F_f = 0\\F_f = F$

So, we can say that the frictional force is equal to the drive force: 205 N.

When the drive force is shut off, only the frictional force is acting on the snowmobile. According to Newton's second law, the frictional force will be equal to the product between the mass of the snowmobile (m) and its acceleration (a):

$-F_f = ma$

and since we know the mass, we can find the acceleration:

$a=-\frac{F_f}{m}=-\frac{205 N}{134 kg}=-1.53 m/s^2$

Where the negative sign means that it is a deceleration, since the frictional force acts in a direction opposite to the motion of the snowmobile.

Now we can finally find the distance covered by the snowmobile before halting by using:

$v^2 - u^2 = 2ad$

where

v = 0 is the final speed

u = 6.36 m/s is the initial speed

a = 1.53 m/s^2 is the acceleration

d = ? is the distance

Solving for d,

$d=\frac{v^2-u^2}{2a}=\frac{0-(6.36 m/s)^2}{2(-1.53 m/s^2)}=13.2 m$

(b) 4.2 s

The acceleration of the snowmobile is related to the time t by:

$a=\frac{v-u}{t}$

Re-arranging for t and putting numbers in, we find the time the snowmobile takes to stop:

$t=\frac{v-u}{a}=\frac{0-(6.36 m/s)}{-1.53 m/s^2}=4.2 s$