# Homework Help: Deceleration of a mass down hill

1. Aug 11, 2010

### Big G

Hi all,

Embarassed to ask as seems simple, i have done the maths but answer seems too low! - have looked all over for worked example to follow but no luck :(

I have a free rolling mass of 2200kg that is being driven down a slope at 0.204m/s.
This mass must be braked to a stop in 0.5m i need to calculate the brake force required to do this.

I have calculated mg sin8.5° = 3190N

v2=u2+2as => 0 = 0.041616 + 2a x 0.5m

a = 2 x -0.041616 /0.5 => cancels out to give 0.041616m/s/s

F=ma so 2200 x 0.041616 = 91.55N

Reason i'm scared is that the force to decelerate (91.55N) seems so small but reletive to the speed 0.5m is quite a long distance i suppose!!

Can anyone confirm my reasoning please!!

Thanks!

2. Aug 12, 2010

### Big G

No one able to confirm or condem my working???

3. Aug 12, 2010

### Staff: Mentor

It's not clear what the problem is that you are trying to solve.

Can you state the problem completely. (Is something rolling down a hill? What's the angle of the hill? What forces act on the object?)

4. Aug 12, 2010

### Big G

Hi,

Sorry, it is a free rolling 'truck' but is being driven at a speed of 0.204m/s - it is driven directly on the tyres and due to the gearbox driving the tyre being 'self sustaining' the truck will not speed up down the hill, only continue at 0.204m/s.

The angle of the slope is 8.5°.

Only forces acting are to be assumed as gravity and the driving force.

At the point in question the power is removed from the gearbox and a short placed across the electric motor therefore using the gearbox to brake the truck to a halt in 0.5m

Hope that clarifys a bit...

Thanks!

5. Aug 12, 2010

### Staff: Mentor

OK, I think I understand what you're doing now. Your work looks correct.

6. Aug 12, 2010

### Big G

Thank you very much for your help,

I think i can trust someone with that many posts to their name!!! :)