Asked by TP

Sam, whose mass is 76kg , takes off across level snow on his jet-powered skis. The skis have a thrust of 220N and a coefficient of kinetic friction on snow of 0.1. Unfortunately, the skis run out of fuel after only 15s .

How far has Sam traveled when he finally coasts to a stop?
Answered by Devron
After the skis run out of fuel, the Force of kinetic friction will do work on the skier.


220N=Fµk*d

Solve for d:

220N=m*g*µk*d

220N=(76kg)*(9.8m/s^2)*(0.1)*d

220N=74.48N*d

d=220N/74.48N

d=2.95m=3m


I could be wrong.
Answered by Devron
Ignore the work done.

This is a multistep problem and Fnet doesn't equal 220N.

Fnet=Fthrust-Fk

Fnet=220N-74.48N=145.52N

Solve for acceleration:

F=ma

a=F/m

a=145.52N/76kg

a=1.91m/s^2

Use the following kinematic equation:

d=vi*t+1/2at^2

where

Vi=0m/s
a=1.91m/s^2
t=15s
and
d=?

Solve for d:

d=0 + 1/2(1.91m/s^2)*(15s)2

d=214.8m

Use the following kinematic equation to find the velocity of the skier:


d=Vf*t-1/2at^2

Where

d=214.8m
Vf=?
a=1.91m/s^2
and
t=15s

Solve for Vf:

Vf=[d+1/2at^2]/t

Vf=[214.8m+1/2(1.91m/s^2)(15s)^2]/15s

Vf=28.59m/s



The first leg of the trip, what about the second leg of the trip?


The frictional force will be equal the net force after the fuel has stopped providing a thrust.

1/2mv^2=Fk*d

1/2mv^2=mk*m*g*d

1/2v^2=mk*g*d

Where

v=28.59
mk=0.1
g=9.8m/s^2
and
d=?

Solve for d:

1/2(28.59m/s)^2=(0.1)(9.8m/s^2)*d

408.69=0.98*d

d=417m

d=417m+215m=632m


*** Still not sure that this completely correct, so hopefully someone comes by and double checks this.



Answered by Anonymous
correct!
Answered by Mama
Anonymous, imagine responding 3 years later, my home boy Devron been waiting for an answer for 3 years
Answered by Anonymous
Lmao so we all just here for mastering physics huh XD
Answered by The procrastinator
Yep, I sorta did what my name says. I'm also really late to the party
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