After the skis run out of fuel, the Force of kinetic friction will do work on the skier.
220N=*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.
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?
6 answers
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.
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.
correct!
Anonymous, imagine responding 3 years later, my home boy Devron been waiting for an answer for 3 years
Lmao so we all just here for mastering physics huh XD
Yep, I sorta did what my name says. I'm also really late to the party
0 answers