Motion on a rough inclined plane

[Taylan]

Homework Statement

A car is moving up a rough inclined plane/road of 10 degrees with a velocity of 10m/s. The coefficient of friction is equal to 0.5 and the mass of the car is 10kg. The car brakes. After what distance does the car stop?

Homework Equations

F=ma
v^2=u^2+2as
u= initial velocity
v= final velocity

The Attempt at a Solution

s=?
u= 10m/s
v= 0m/s
a= ?

a= net force/ mass

= [(-0,5 x 10kg x 9.81m/s2 x cos10) - 10kg(9.81m/s2)(sin10) ]/ 10kg = -6.53m/s2

and using v^2=u^2+2as , s is found to be 7.65m

What I am confused about is the net force for which I assumed to be:
(-0,5 x 10kg x 9.81m/s2 x cos10) - 10kg(9.81m/s2)(sin10)

So I just calculated -[ the weight of the car acting downwards + friction ( found by μR) ]. The car is actually moving upwards. But there is no force by the engine in upwards direction? So is the net force really acting downwards?

 

[scottdave]

Since both force components are acting to slow the car down, should you be subtracting or adding them?

 

[Taylan]

Since both force components are acting to slow the car down, should you be subtracting or adding them?

I guess I can either go for -(F1+F2) or -F1-F2 . I applied the second one

 

[scottdave]

I guess I can either go for -(F1+F2) or -F1-F2 . I applied the second one

OK, I missed the minus sign before the 0.5, the first time I looked at it. So you had it correct.

 

[Taylan]

Alright thanks a lot! so there isn't any force acting upwards by the engine since the car is braking right? like although the car is moving upwards, there are only forces that are acting downwards to be considered which are; friction and mg ?

 

[haruspex]

Alright thanks a lot! so there isn't any force acting upwards by the engine since the car is braking right? like although the car is moving upwards, there are only forces that are acting downwards to be considered which are; friction and mg ?

Yes.