Question

A bus of mass m kg is drawn up an inclined road of inclination θ at a constant acceleration a m/s2. If the coefficient of friction is μ, the work done by the engine in 10s starting from rest is:

Pls answer fast

Answer 1

Given: mass of bus = m kg

           The angle of inclination = θ

           acceleration = a ms⁻²

          coefficient of friction =  μ

          Time = 10 seconds

To find: Work done by the engine

Solution:

• A bus is descending an inclined plane and is only affected by the force of gravity. The friction force is the non-conservative force, it causes hindrance to the movement.

• Acceleration of the bus due to gravity on an inclined plane of angle θ = gSinθ, and when the friction comes into play then acceleration becomes = μgSinθ.

Distance travelled by bus in 10 seconds

S = ut + 1/2at²

a =  μgSinθ( acceleration)

S = 0 + 1/2(μgSinθ)×10²

 S  = 50 μgSinθ

Now, lets calculate the driving force,

Driving force = mgSinθ - μmgCosθ

Work done by the engine is the product of the driving force and distance it cover.

Work done = F×S

                  = (mgSinθ - μmgCosθ) 50 μgSinθ

                 = 50μgSinθ(mgSinθ - μgCosθ)

Therefore, the work done by the engine is 50μgSinθ(mgSinθ - μgCosθ)

Work done by the bus will be more whenfriction comes into play.