A car of mass 2 320 kg is descending a sloping road, inclined at 10.0° to the horizontal. The driver sees a hazard and brakes to make an emergency stop. The distance travelled while braking is 22.5 m. The initial speed of the car was 13.4 m s−1. Calculate the average braking force. Assume air resistance is negligible.

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.

0.19 m/s and 0.45 m/s

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y