Observation
Question

Move a 1-kg block from smooth to rough surface. Observe how work done by frictional force is related to change in the kinetic energy of the block.

Rough surface
Smooth surface
Kinetic energy
Work done by friction
Change in K.E.
2 m/s 3 m/s 4 m/s 5 m/s
Velocity of the block
2 N 3 N 4 N 5 N
Frictional force

What will be the work done (in J) by a car engine to change the velocity of the car of mass m from 5 m/s to 10 m/s?


\(\frac{25}{2} \times m\)
\(\frac{50}{2} \times m\)
\(\frac{75}{2} \times m\)
\(\frac{100}{2} \times m\)

Solution

•  The work done is equal to the change      in the kinetic energy.

•  The velocity of the car changes from          5 m/s to 10 m/s.

•  The kinetic energy of the car changes     from 1/2 × m × 5² to 1/2 × m × 10².

•  The work done by the engine will be     change in K.E i.e.
    1/2 × m × (10² - 5²) = 75/2 × m.

You just explored that according to the work-energy theorem, the work done is equal to the change in the kinetic energy.