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.

when jerome paddle in flowing river towards east the net speed of jerome will increase

so the net speed of jerome will be given as

now the speed of paul will be same as it is given as he is running on the ground

so

so here we can say since the speed of jerome is more so he will win the race

now as we know the distance so the time difference is given as

so jerome will reach first and the time difference will be 203.55 s or 3.4 min