Physics question

[golden]

How would I do this question? The answer (which is in the back of the book): 7.12 × 10^3 N

I AM ONLY REFERRING TO PART G.

I keep getting 6.0 x 10^3 N.

Here's my working:
The change in kinetic energy is the work done:
1/2mv^2 = 5.4 x 10^5 J
W = Fxcos(theta)
W = F(90)cos5.74
W = 5.4 x 10^5 N m
F = 6.0 x 10^3 N.

I'm pretty sure that 6.0 x 10^3 is not the answer.

Thanks.

[echenzi]

Hmm.. There may be an error in your book..

Using the kinetics equation v^2= u^2 + 2as, you are able to calculate a value for a.

and simple F=ma to find the magnitude of the force...

In this case... v= 0 U= 30, s = 90, m= 1200kg

you get 0= 900+ 180a.

-900= 180a

a= -5.

|F= ma|

= 6000N.

I may be wrong!!!



F=ma

[dptjandra]

Definitely 7.12 x 10^3 N?  Seems a strange number of sig figs.  Though I can see where they might've pulled 7.2 x 10^3 N from.

[QuantumJG]

Do you mind typing out ALL of g? The question is partially cutoff

Ok here is why you got 6kN instead of 7.2kN (which is the right answer).

So what you were doing is correct, the NET acceleration is -5ms^-2. The question wants the force applied to the car which must take into account the acceleration due to gravity which is 1ms^-2 (due to the 1 in 10 incline and assuming g = 10ms^-2), so the car's acceleration due to the brakes is -6ms^-2 implying the magnitude of the force applied to it is 7.2kN.

[golden]

Do you mind typing out ALL of g? The question is partially cutoff

Ok here is why you got 6kN instead of 7.2kN (which is the right answer).

So what you were doing is correct, the NET acceleration is -5ms^-2. The question wants the force applied to the car which must take into account the acceleration due to gravity which is 1ms^-2 (due to the 1 in 10 incline and assuming g = 10ms^-2), so the car's acceleration due to the brakes is -6ms^-2 implying the magnitude of the force applied to it is 7.2kN.

I see it now thanks. It may seem strange for some people that you first have to figure out the force as if it were not inclined and use that to figure out the acceleration. Then you find the acceleration due to the incline, add that on and find the braking force on the incline.

[dptjandra]

If that method seems strange for you, it is equivalent to find the total energy which must be "lost" (ie. changed to heat by the brakes) as the car moves from its original position to its final resting position...

That is: KE + increase in GPE = 1/2mv^2 + mgh = 540000 + (1200)(9.(9) = 645840 W
We use W = F.x (consider that the car's center of mass moves in a path that is parallel to the incline of the plane at all times)
So F x 90 = 645840
F = 7176 N = 7.2 kN (same answer)