f = 1.32×10^4 N
Explanation:
We can use the work-energy theorem to find the work done by the braking force f:
W = ∆KE + ∆PE
= (KEf - KEi) + (PEf - PEi)
= [(1/2)mvf^2 - (1/2)mvi^2] + (mhf - mghi)
At the bottom of the slope, vf = 0 and hf = 0 and hi = dsin10° (d = braking distance) so work W becomes
W = -[(1/2)mvi^2 + mgdsin10°]
= -m[(1/2)vi^2 + gdsin10°]
= -(2320kg)[(1/2)(13.4m/s)^2 + (9.8 m/s^2)(22.5m)sin10]
= -2.97×10^5 J
Since W = fd, where f is the braking force, we can now solve for f:
f = W/d = (-2.97×10^5 J)/(22.5 m)
= -1.32×10^4 N
Note: the negative sign means that it is a dissipative force.
