# A 1500kg car is coasting on a horizontal road with a speed of 18 m/s . ?

After passing over an unpaved, sandy stretch 25.0 m long, car's speed has decreased to 14 m/s . Find the magnitude of the average net force on the car in the sandy section of the road.

Relevance
• To slow down the car goes through a deceleration.

We have u = 18, v = 14 and s = 25

Then, using v² = u² + 2as we get:

14² = 18² + 2a(25)

so, a = (14² - 18²)/50

Hence, a = -2.56 ms⁻²

Now, F = ma

Then, F = 1500(-2.56) => -3840 Newtons

Note, work done is force x distance

And work done is change in kinetic energy

so, F(25) = (1/2)(1500)(14² - 18²)

Hence, F = (1/2)(60)(14² - 18²)

i.e. F = 30(14² - 18²) => -3840 N

:)>

• ΔKE = m/2 (14^2-18^2) = 750*-128 = -96,000 joule = F*d

friction force F = Δ KE /d = -96/25 = -3.84 kN

• average speed on the sandy section

v_av = (18+14)/2 = 32/2 = 16 m/s

time spent on sandy section

Δt = s / v_av = 25 / 16 = 1.56 s

average deceleration

a = Δv/Δt

Δv = speed after - speed before

a = (14-18) / 1.56 = -4 / 1.56 = -2.56 m/s²

the magnitude of the average net force

F = m a

F = 1500 * (-2.56) = -3840N

"-" means it's a braking force