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A car of mass 1700kg moves in a straight line along a slope that is at an angle θ to the
horizontal, as shown in Fig. 3.1.
The car moves at constant velocity for a distance of 25m from point A to point B.
Air resistance and friction provide a total resistive force of 440N that opposes the motion of the car.
For the movement of the car from A to B:
(i) state the change in the kinetic energy.
[How do you find the change in kinetic energy?..]

A car of mass 1700kg moves in a straight line along a slope that is at an angle θ to the horizontal as shown in Fig 31 The car moves at constant velocity for a class=

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oladayoademosu

The change in the kinetic energy is (22000 + 416500sinθ) J

Net force on car

The net force on the car F' = F - f - N where

  • F = force due to car,
  • f = total resistive force = 440 N and
  • N = horizontal component of weight of car = mgsinФ where
  • m = mass of car = 1700 kg,
  • g = acceleration due to gravity = 9.8 m/s² and
  • θ = angle of slope

So, F' = F - f - N

F' = F - f - mgsinθ

Since the car moves at constant velocity, the net force F' = 0

So, F' = F - f - mgsinθ

F - f - mgsinθ = 0

F = f + mgsinθ

Work-kinetic energy theorem

From the work-kinetic energy theorem,

The kinetic energy change of the car ΔK equals the work done by force on car, W

ΔK = W = Fd where

  • F = force on car and
  • d = distance moved by car = 25 m

ΔK = Fd

= (f + mgsinθ)d

=  fd + mgdsinθ

Change in kinetic energy of car

Substituting the values of the variables into the equation, we have

ΔK = fd + mgdsinθ

= 440 N × 25 m + 1700 kg × 9.8 m/s² × 25 m × sinθ

= 22000 Nm + 416500sinθ Nm

= (22000 + 416500sinθ) J

So, the change in the kinetic energy is (22000 + 416500sinθ) J

Learn more about change in kinetic energy here:

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