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.
How far has Sam traveled when he finally coasts to a stop?
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.
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.
The thrust force from the jet-powered skis can be calculated using Newton's second law, which states that the force is equal to the mass multiplied by the acceleration (F = ma). Rearranging the formula, we can calculate the acceleration (a) by dividing the thrust force (F) by the mass (m):
a = F / m
Plugging in the values, we have:
F = 220 N (thrust force)
m = 76 kg (mass of Sam)
a = 220 N / 76 kg
a ≈ 2.89 m/s²
Now that we have the acceleration, we can determine the distance traveled by using the formula for distance covered with constant acceleration:
d = v₀t + 0.5at²
We know that the skis run out of fuel after 15 seconds, so the time (t) is 15 seconds. To find the initial velocity (vâ‚€), we can use the fact that Sam starts from rest, so vâ‚€ = 0.
Plugging in the values, we have:
vâ‚€ = 0 m/s
t = 15 s
a = 2.89 m/s²
d = (0 m/s)(15 s) + 0.5(2.89 m/s²)(15 s)²
d = 0 + 0.5(2.89 m/s²)(225 s²)
d = 0 + 329.175 m
d ≈ 329.175 m
Therefore, Sam has traveled approximately 329.175 meters when he finally coasts to a stop.