After a day of testing race cars, you decide to take your own 1550-kg car onto the test track. While moving down the track at 10.0 m/s, you uniformly accelerate to 30.0 m/s in 10.0 s. What is the average net force that the track has applied to the car during the 10.0-s interval?
2006-10-23 14:40:03 · 3 answers · asked by yoo whoo 1 in Science & Mathematics ➔ Physics
Vf = Vi + a t
Where a = acceleration, t = 10 seconds, Vf = 30 m/s, Vi = 10 m/s
solve for the acceleration.
F = m a
Where F is the force exerted by the track, m = 1550 kg, and a is the acceleration you found above.
Solve for F. (the turning wheels push back on the track and the track pushes forward on the car).
Aloha
P.S. This is a subtlety, the Normal force (mentioned in the answer above) doesn't enter into the problem because it is cancelled by weight and there is no net vertical force on the car. Don't be confused by this... I mention it more for Moo's benefit. And spoilers, etc. would only appear in an Engineering context.
2006-10-23 14:46:14 · answer #1 · answered by Anonymous · 0⤊ 0⤋
Calculate the average acceleration,a, using the formula:
v=u+at
where v=final velocity, u the initial velocity, a the acceleration, and t the time.
Substitute:
30=10+a*10
30-10=10a
20=10a
a=20/10
=2m/s^2
The net force is given by the formula F=Ma, where F equals the net force, M the mass of the car, and a the acceleration. Substitute known values:
F=1550*2
=3100N
2006-10-24 05:22:08 · answer #2 · answered by tul b 3 · 0⤊ 0⤋
track has applied to the car.....u mean the normal force?
if so then its just weight times gravity
1550kg ( 9.81 m/s^2)
but if this is higher level physics, ur gonna have to include the downforce caused by air pressure. and that would involve the spoiler and body shape of that car.
2006-10-23 14:44:36 · answer #3 · answered by Mr.Moo 4 · 0⤊ 0⤋