Deceleration of a mass down hill

[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

Add together gives 3281.555N

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!

 

[Big G]

No one able to confirm or condem my working?

 

[Doc Al]

No one able to confirm or condem my working?

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?)

 

[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!

 

[Doc Al]

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

 

[Big G]

Thank you very much for your help,

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