Question

A 6-kg toboggan is kicked on a frozen pond, such that it acquires a speed of 1.6 m/s to the right. The coefficient of friction between the pond and the toboggan is 0.18. Determine the distance that the toboggan slides before coming to rest. Answer: __________ m (round to the nearest hundredth)

Answer 1

.73 m

Step-by-step explanation:

Kinetic Energy = 1/2 m v^2 = 1/2 * 6 * (1.6)^2 = 7.68 J

The work of friction must equal this for the sled to stop

Force of friction = .18 * 6 kg * 9.81 m/s^2 = 10.59 N

Force * Distance = work

10.59 * d = 7.68 J

d = .73 m

Answer 2

Since the toboggan is sliding through the horizontal surface of the frozen pond, its acceleration is given by:

`a=\mu g`

The distance that an accelerating object travels while decelerating from a speed v to a stop with an acceleration a is:

`d=(v^2)/(2a)`

Replace a=μg as well as v=1.6m/s, g=9.8m/s^2 and μ=0.18:

`d=((1.6(m)/(s))^2)/(2(0.18)(9.8(m)/(s)))=0.7256...m\approx0.73m`

Therefore, to the nearest hundredth, the distance traveled by the toboggan is 0.73m.