Finding the Braking Force of a Truck on an Inclined Plane

[Bill_Nye_Fan]

Homework Statement

A truck of mass 1.5 tonnes (1500 kg) is moving at a speed of 36 km/h down a hill of slope 1 in 6. The driver applies the breaks and the truck comes to rest after 2 seconds. Find the breaking force, which is assumed to be constant, ignoring any other resistant forces.

Homework Equations

$v=u+at$
A slope of 1 in 6 means $\sin \left(\theta \right)=\frac{1}{6}$

The Attempt at a Solution

First I converted 36 km/h into 10 m/s.
I then rearranged and solved $v=u+at$ to get $a=\frac{v-u}{t}=\frac{0-10}{2}=-5\ ms^{-2}$

This makes sense as the negative sign is denoting that the breaking force is acting in the opposite direction to motion. I noted that this deceleration is being caused purely by the breaks, therefore the breaking force should be $F_b=-5\cdot 1500=-7500$ however this isn't right. I know I didn't use the slope of 1 in 6 part, but I don't understand why I would need it.

 

[TSny]

Draw a free body diagram and consider the net force acting along the slope.

 

[kuruman]

I know I didn't use the slope of 1 in 6 part, but I don't understand why I would need it.

What if the truck were braking while moving uphill or along a horizontal surface? According to your solution the braking force will be the same in all cases. Does this make sense?

 

[Dr.D]

I think you really want the brake force, not the break force.