Answer:
The force required by the brakes is [tex]3.67\times 10^{3} N[/tex]
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
[tex]g = 10 m/s^{2}[/tex]
Force required by the brakes to stop the truck, [tex]F_{b}[/tex] can be calculated by using the law of conservation of energy.
Now,
Kinetic Energy(K.E) downhil, K.E = reduction in potential energy, [tex]\Delta PE[/tex]
[tex]\Delta PE = Mgh = 2200\times 10\times 50 = 1100 kJ[/tex]
Work done is provided by the decrease in K.E,i.e., change in potential energy.
W = [tex]F\times x = 300 F = 1100\times 10^{3}[/tex]
F = [tex]3.67\times 10^{3} N[/tex]
