Sam, whose mass is 72 kg , takes off across level snow on his jet-powered skis. The skis have a thrust of 190 N and a coefficient of kinetic friction on snow of 0.1. Unfortunately, the skis run out of fuel after only 9.0 s. How far has Sam traveled when he finally coasts to a stop?

Given Information:

Mass = 72 kg

Force exerted by ski = 190 N

coefficient of kinetic friction = 0.1

ski runs out of fuel after = 9 sec

Required Information:

Distance traveled when he stops = ?

d = 179.33 m

Explanation:

The problem can be divided into two parts

Part 1: before running out of fuel

In the first part, the forces acting on Sam are

Ski thrust and Force of friction

Force of friction = kmg

where k is the coefficient of kinetic friction, m is Sam's mass and g is gravity 9.8 m/s²

Force of friction = 0.1*72*9.8

Force of friction = 70.56 N

so the net force acting on Sam is

Fnet = ski thrust - force of friction

Fnet = 190 - 70.56

Fnet = 119.44 N

According to Newton's second law of motion

F = ma

a = F/m

a = 119.44/72

a = 1.65 m/s²

The distance traveled can be found using kinematic equation

d = vi + 0.5at²

d = 0 + 0.5*1.65*9²

d = 66.82 m

The speed at this point is

vf = (d + 0.5at²)/t

vf = (66.82 + 0.5(1.65)(9)²)/9

vf = 14.85 m/s

Part 2: after running out of fuel

When the fuel runs out the ski is no longer applying any force so the only force acting on Sam is force of friction

kmg = ma

kg = a

a = 0.1*-9.8

a = -0.98 (minus sign due to deceleration)

The distance traveled can be found using kinematic equation

2ad = vf² - vi²

d = (vf² - vi²)/2a

d = (vf² - vi²)/2a

d = (0² - 14.85²)/2*-0.98

d = 112.51 m

How far has Sam traveled when he finally coasts to a stop?

We have to sum the distance before and after Sam runs out of fuel because he stops after covering the sum of these distance

d = 66.82 + 112.51

d = 179.33 m

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j