Question

The engine of a car moving with a speed 36 km per hour is switched off as it reaches a inclined road, inclined at 30 degree with the horizontal if the coefficient of friction between the road surface And The Wheels of the car is 1\√3 find the distance travelled by the car on the inclined Road before it comes to rest.

Answer 1

Initial energy = kinetic energy = KE = mv^2 /2

m = mass of the car

v = initial velocity = 36km/hour = 10 m/sec

Angle of inclination θ = 30 degrees

Work done by friction = Ffric * d=μ(mg)cosθd

where d = distance traveled up the incline

μ – Coefficient of friction = 1/√3

At the top of the incline, the car only has potential energy:

PE = mgh = mg(dsinθ)

KE – Efric = PE

½mv^2 − μ(mg)cosθd = mg(dsinθ)

mass cancels in every term giving:

v^2/2− μ*g*cosθ*d = g*(d*sinθ)

(10 * 10 / 2) – ((d * 9.8 * cos*30)/√3) = 9.8 * d * sin(30)

50 – 4.9*d = 4.9*d

D = 10.2 meters

Car goes 10.2 meters up and stops.

(starts moving down, if breaks are not applied)