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## 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

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?

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Physics
1 year
2021-08-30T14:21:55+00:00
2021-08-30T14:21:55+00:00 2 Answers
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## Answers ( )

Answer:181.11 mExplanation:Net Forceis the sum of all theresultant forceof 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;[tex]F_{net} = F_{applied} – F_{friction} \\[/tex] ……………………….1

but [tex]F_{friction}[/tex] = 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/[tex]s^{2}[/tex]

Substituting into equation 2 we have;

[tex]F_{friction}[/tex] = 0.1 x 72 x 9.81

[tex]F_{friction}[/tex] = 70.63 N

Now we substitute our answer in equation 1;

[tex]F_{net} = 190 N – 70.63 N \\[/tex]

[tex]F_{net}[/tex] = 119.37 N

We have to calculate the net acceleration in order to get our velocity .[tex]F_{net}[/tex] = m[tex]a_{net}[/tex]

[tex]a_{net}[/tex] = [tex]\frac{F_{net}}{m}[/tex]

[tex]a_{net}[/tex] = [tex]\frac{119.37 N}{72 kg }[/tex]

[tex]a_{net}[/tex] = 1.66 [tex]m/s^{2}[/tex]

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 [tex]m/s^{2}[/tex]

t is the time which is 9 s

substituting the values in equation 3 we have

v = 1.66 [tex]m/s^{2}[/tex] 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;[tex]S_{1} = ut +\frac{1}{2}at^{2}[/tex]

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

[tex]S_{1} = \frac{1}{2} *1.66 m/s^{2} *9^{2}[/tex]

[tex]S_{1}[/tex]

= 67.23 mAlso we have to calculate Sam’s distance after nine seconds using equations of motion express below;[tex]v^{2} -u^{2} = 2as[/tex]

making S the subject formula we have;

[tex]S_{2} =\frac{v^{2}-u^{2} }{2a}[/tex] ………………………………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/[tex]s^{2}[/tex] = 0.98 m/[tex]s^{2}[/tex]

We substitute our values into equation 4 to get our remaining distance;[tex]S_{2} = \frac{(14.94 m/s)^{2} }{2(0.98 m/s^{2} )}[/tex]

[tex]S_{2}[/tex]

= 113.88 mTherefore the total distance S = [tex]S_{1}[/tex] + [tex]S_{2}[/tex]

S = 67.23 m + 113.88 mThe total distance covered by Sam is181.11 mGiven 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 fuelIn 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 fuelWhen 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