Question

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?

Answer 1

Answer:

181.11 m

Explanation:

Net Force is the sum of all the resultant force of all the forces acting on a body. The net force experienced by the jet is the difference of the Applied force and frictional force and this is represented in equation 1;

$F_{net} = F_{applied} - F_{friction}$ ............................1

but $F_{friction}$ = μmg .......................................2

Where μ is the coefficient of friction = 0.1

          m is the mass of the body = 72 kg

          g is the acceleration due to gravity 9.81 m/$s^{2}$

Substituting into equation 2 we have;

$F_{friction}$ = 0.1 x 72 x 9.81

$F_{friction}$ = 70.63 N

Now we substitute our answer in equation 1;

$F_{net} = 190 N - 70.63 N$

$F_{net}$ = 119.37 N

We have to calculate the net acceleration in order to get our velocity .

$F_{net}$ = m$a_{net}$

$a_{net}$ = $\frac{F_{net}}{m}$

$a_{net}$  = $\frac{119.37 N}{72 kg }$

$a_{net}$  = 1.66 $m/s^{2}$

Speed can be express as;

v = u + at ...................3

where u is the initial velocity, which is 0 in this case.

a is the net acceleration  = 1.66 $m/s^{2}$

t is the time which is 9 s

substituting the values in equation 3 we have

v =  1.66 $m/s^{2}$ x 9 s

v = 14.94 m/s

Calculating for Sam's distance for the first 9 seconds, using the equation of motion we have;

$S_{1} = ut +\frac{1}{2}at^{2}$

the jet was initially at rest so initial velocity is 0  and $a_{net}$  = 1.66 $m/s^{2}$

$S_{1} = \frac{1}{2} *1.66 m/s^{2} *9^{2}$

$S_{1}$ = 67.23 m

Also we have to calculate Sam's distance after nine seconds using equations of motion express below;

$v^{2} -u^{2} = 2as$

making S the subject formula we have;

$S_{2} =\frac{v^{2}-u^{2} }{2a}$ ....................................4

v is the maximum velocity after the fuel finished  and its 14.94 m/s and a is the acceleration along the horizontal plane which put into consideration the coefficient of friction. a = μg = 0.1*9.8 m/$s^{2}$ = 0.98 m/$s^{2}$

We substitute our values into equation 4 to get our remaining distance;

$S_{2} = \frac{(14.94 m/s)^{2} }{2(0.98 m/s^{2} )}$

$S_{2}$ = 113.88 m

Therefore the total distance S = $S_{1}$ + $S_{2}$

S = 67.23 m + 113.88 m

The total distance covered by Sam is 181.11 m

Answer 2

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 = ?

Answer:

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