Question

Sam, whose mass is 71kg , takes off across level snow on his jet-powered skis. The skis have a thrust of 250N and a coefficient of kinetic friction on snow of 0.1. Unfortunately, the skis run out of fuel after only 10s.What is Sam's top speed? (Express Answer in 2 sig figures & correct units)How far has Sam traveled when he finally coasts to a stop? (Express Answer in 2 sig figures & correct units)

Answer 1

Answer:

Sam's top speed = 25 m/s

Distance Sam travels before coastring to a stop = 360 m

Explanation:

The thrust from the skis = 250 N

Frictional force on Sam = coefficient of friction × Sam's weight

                                       = (0.1)(71 kg)(9.8 m/s²) = 69.58 N

The effective force moving Sam = 250 N - 69.58 N = 180.42 N

The acceleration of Sam due to this force = 180.42 N ÷ 71 kg = 2.54 m/s²

Sam starts from rest. Initial velocity, u = 0 m/s. The skis run out after time, t = 10 s.

Using the equation of motion,

v = u + at

v = 0 + (2.54 m/s²)(10 s) = 25.4 m/s ≈ 25 m/s

This is Sam's top speed.

v² = u² + 2as

$s = \dfrac{v^2-u^2}{2a} = \dfrac{(25.4\text{ m/s})^2-(0\text{ m/s})^2}{2\times(9.8\text{ m/s}^2)} = 32.92\text{ m}$

The distance travelled while the skis were working is determined using the equation of motion

For the stopping motion, his initial velocity is 25.4 m/s and final velocity = 0 m/s. His acceleration is determined by the frictional force since there is no thrust from the skis.

a = -69.58 N ÷ 71 kg = -0.98 m/s²

(It is negative because it acts in opposite direction to the velocity)

Using the equation of motion

v² = u² + 2as

$s = \dfrac{v^2-u^2}{2a}$

$s = \dfrac{(0\text{ m/s})^2-(25.4\text{ m/s})^2}{2\times(-0.98\text{ m/s}^2)} = 329.16\text{ m}$

Total distance travelled = 32.92 m + 329.16 m = 362.08 m ≈ 360 m