- #1
Bill_Nye_Fan
- 31
- 2
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.