Question

1.a car moves at a constant velocity up an inclined plane angle of 28 degrees with the horizontal.The coefficient of friction is 0.387. The force exerted by the engine on the car is 588N...

A.Calculate the of the car?
B. Frictional force acting on the car

Answer 1

Yes it is physical science

Answer 2

First step: draw a free body diagram, showing
1. W=mg
2. down-plane component: mg sin(θ)
3. normal component, N: mg cos(θ)
4. friction (parallel to plane, downwards) = μN
5. applied force, F=588 N

Since car is moving up the plane at constant velocity, forces along the plane are in equilibrium, hence
Σ down-plane forces Σ up-plane forces,
or
mg sin(θ)+μN = F
(A)Solve for m (mass of the car).
(B)frictional force actig on the car = μN

If you do not know how to draw a free-body diagram, read:
http://www.physicsclassroom.com/class/vectors/Lesson-3/Inclined-Planes

Answer 3

Thanks so my mass is 157.62kg? Am I correct?

Answer 4

I get about half of that, which is pretty small for a car. A small MG-midget weighs more than that, unless it is an ATV.

Can you check if you have a factor of 2 missing somewhere? If you don't spot it, tell me what you have for N (normal component), the value of g that you use.

Answer 5

I don't have a factor of two for g I used 9.8?

Answer 6

What did you get for N?

Answer 7

I have
m(9.8)sin(28)+m(9.8)cos(28)*0.387=588
from which
m=588/(9.8(sin(28)+0.387cos(28))
Do you have the same expression?

Answer 8

I have this m(9.8)(sin28)+(0.387)m(9.8)(cos28)=588

My mass is 73.96kg just realized now I made an error while pressing the values on my calculator...... Thanks alot

Answer 9

You're welcome! :)