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From Mechanics

Theory

Exercises

Video

Key Points

The following theory is for a particle of mass m, having a speed v and at a height h above the ground.

Kinetic Energy:  21mv2

Potential Energy:  mgh

Total energy is the sum of the potential and kinetic energy of the particle.

Energy lost due to friction = Work done against friction

If F is a constant friction force and d is the distance the particle moves,

Work done against friction: Fd

Thus if there is no friction or friction is negligible the particle does not lose energy. In other words its energy is constant (conserved).

In the above, any force which resists motion is to be regarded as a frictional force.

Example 17.1

A ball, of mass 300 grams, is hit from ground level, so that its initial speed is 16 ms−1. At its maximum height the ball travels at 6 ms−1.

a) Find the maximum and minimum kinetic energy of the ball.

b) If the ball had been projected vertically upwards what would be the minimum kinetic energy of the ball?

Solution

a) The kinetic energy is at a maximum when the velocity is 16 ms−1.

KE=210.3162=38.4 J

The kinetic energy is at a minimum when the velocity is 6 ms−1.

KE=210.362=5.4 J

b) If the ball was projected vertically, it would have a velocity of zero at its maximum height and hence the minimum kinetic energy would be zero.

Example 17.2

A ball, of mass 20 grams, hits a wall travelling at 4 ms−1 and rebounds at 3 ms−1. Find the energy lost during the bounce.

Solution

Find the kinetic energy before the bounce:

KE=210.0242=0.16 J

Find the kinetic energy after the bounce:

KE=210.0232=0.09 J

The loss of energy can then be calculated:

Energy Lost =0.16−0.09=0.07 J

Example 17.3

A roller coaster has carriages which are modelled as particles of mass 500 kg. A carriage starts at A with speed 12 ms−1. It moves, as shown in the diagram, through the points B, C and D. Find the kinetic energy and speed of the carriage at each of these points.

Solution

Find the total energy of the carriage at A.

Total Energy=mgh+21mv2=5009.81+21500122=4900+36000=40900 J

At the point B:

40900=5009.85+KEKE=40900−24500=16400 J

16400=21500v2v=25016400=8.1 ms-1 

At the point C there is no potential energy, so the kinetic energy is 40900 J. The speed can then be calculated.

40900=21500v2v=25040900=12.8 ms-1 

At the point D:

40900=5009.83+KEKE=40900−14700=26200 J

26200=21500v2v=25026200=10.2 ms-1 

Example 17.4

A bungee jumper has mass 60 kg and uses a rope of length 40m. The jumper comes to rest after falling a distance of 60 m.

a) Find the speed of the jumper when the rope first becomes taut.

b) Find the elastic potential energy stored in the rope, when it is fully stretched.

Solution

a) When the rope becomes taut the bungee jumper has fallen 40 m. The potential energy lost will be equal to the kinetic energy.

609.840=2160v2v=3023520=28 ms-1 

b) At the lowest point all of the potential energy lost will be equal to the elastic potential energy.

EPE =609.860=35280 J