Question

You may have noticed runaway truck lanes while driving in the mountains. These gravel-filled lanes are designed to stop trucks that have lost their brakes on mountain grades. Typically, such a lane is horizontal (if possible) and about 34.5 m34.5 m long. Think of the ground as exerting a frictional drag force on the truck. A truck enters a typical runaway lane with a speed of 54.0 mph54.0 mph ( 24.1 m/s24.1 m/s ). Use the work-energy theorem to find the minimum coefficient of kinetic friction between the truck and the lane to be able to stop the truck.

Answer 1

Answer:0.85

Explanation:

Given

length of lane=34.5 m

speed of truck $=54 mph \approx 24.1 m/s$

using Work Energy theorem

change in kinetic Energy of truck =Work done by all the forces

Change in kinetic Energy is $=\frac{mv^2}{2}-0$

Work done by Frictional force

$W=f_r\times d$

thus

$\frac{m\times 24.1^2}{2}=\mu m\times g$

$\mu =\frac{24.1^2}{2\times 9.8\times 34.5}$

$\mu =0.85$

Answer 2

Answer:

The coefficient of kinetic friction is found to be 0.85

Explanation:

The work-energy theorem in this case states that:

Loss in Kinetic Energy of the car = Work Done by the Friction

(1/2)mv² = (f)(d)

where,

m = mass of the car

v = speed of the car = 24.1 m/s

d = distance of the lane = 34.5 m

f = force of friction = μR

Therefore,

(1/2)mv² = (μR)(d)

where,

μ = coefficient of kinetic friction between truck and lane = ?

R = Normal Reaction = Weight of truck = mg

Therefore,

(1/2)mv² = (μmg)(d)

(1/2)v² = (μg)(d)

μ = v²/2gd

μ = (24.1 m/s)²/2(9.8 m/s²)(34.5 m)

μ = 0.85