Question

A 2200 kg truck is coming down a hill 50 m high towards a stop sign. What force will the brakes need to provide (in N) in order to stop the truck in the 300 m before the sign? (This is a conservation of energy problem). Use g = 10 m/s^2.

Answer 1

The force required by the brakes is
`3.67* 10^(3) N`

Solution:

As per the question:

Mass of the truck, M = 2200 kg

Height, h = 50 m

distance moved by the truck before stopping, x = 300 m

`g = 10 m/s^(2)`

Force required by the brakes to stop the truck,
`F_(b)` can be calculated by using the law of conservation of energy.

Now,

Kinetic Energy(K.E) downhil, K.E = reduction in potential energy,
`\Delta PE`

`\Delta PE = Mgh = 2200* 10* 50 = 1100 kJ`

Work done is provided by the decrease in K.E,i.e., change in potential energy.

W =
`F* x = 300 F = 1100* 10^(3)`

F =
`3.67* 10^(3) N`