Asked by tom


Question:
You live at the top of a steep (a slope of ???? degrees above the horizontal) hill and must park your 2200 kg car on the street at night.

a) You unwisely leave your car out of gear one night and your handbrake fails.

Assuming no significant frictional forces are acting on the car, how quickly will it accelerate down the hill?

b) You resolve to always leave your car in gear when parked on a slope. If the rolling frictional force caused by leaving the drive--?train connected to the wheels is 5000 N, at what rate will your car accelerate down the hill if the handbrake fails again?

*It was written on my homeowork that the answer for a=2.5m/s^2 and b=0.3m/s^2 but I don't know how they got that answer.
Answered by tom
*It was written on my homeowork that the answer for a=2.5m/s^2 and b=0.3m/s^2 but I don't know how they got that answer.
Answered by Chanz
a) I don't know how they got that answer either.
depends on ????
mg sin???? = ma
b) mg sin????- Ff = ma
Answered by Holly
the sin of the angle is 15 degrees. I just had the exact same homework problem and got the correct value for acceleration by doing what he said above.
Answered by Holly
adding to what I just posted, so the Fnet is = Force of the rolling car (mgsin(15))-Force of friction.
So Fnet=5580N-5000N=580N
And Fnet=ma
So acceleration=Fnet/mass
580N/2200kg= 0.264m/s^2
Answered by Isha
Why is it mgSin15 and not mgtan15 please?
Answered by rose
mgsin15=ma
2200x10xsin15=5694=ma
5694=2200a
a=2.5m/s
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