## Force and Power

One of the basic equations of physics is or If the work is done in a time then we can divide both sides by to obtain or where and We can generalise this equation: If a body is travelling at a velocity against a force then the power needed to overcome the force is It should be noted that at any instant the body may be accelerating or decelerating, so that v is an instantaneous, not necessarily a constant, speed. If a force is applied greater than the force needed to over come the resistive force then the body may accelerate, or the potential energy of the body may increase in some way. Given a power and a velocity we may find the driving force also using Example

A lorry of mass 1500 kg moves along a straight horizontal road. The resistance to the motion of the

lorry has magnitude 750 N and the lorry’s engine is working at a rate of 36 kW.

a) Find the acceleration of the lorry when its speed is 20 m/s.

The lorry comes to a hill inclined at an angle to the horizontal, where The magnitude of the resistance to motion from non-gravitational forces remains 750 N.

The lorry moves up the hill at a constant speed of 20 m/s.

b) Find the rate at which the lorry's engine is now working.

a)  The resultant force=1800-750=1050N

Now use for the lorry – note that here is the resultant force and not the 750N force in the question. b) The lorry does work to overcome the force of friction and to increase the gravitational potenital energy by travelling up the hill. In 1 second the lorry travels a distance v up the hill so from the diagram above, rises a vertical distance    