A forklift uses 5200 J of energy to lift a 50.0 kg mass to a height of 4.0 m at a constant speed. What is the efficiency of the forklift?

Answer 1

The efficiency of a device is given by the ratio of useful output energy to the input energy.

In this case, the useful output energy is the work done to lift the mass to a height of 4.0 m, which is given by the product of the force exerted by the forklift and the distance over which the force is applied:

Work = force x distance

The force exerted by the forklift can be calculated using the equation for work:

Work = force x distance

Solving for force:

force = Work / distance

The work done to lift the mass to a height of 4.0 m is equal to the gravitational potential energy gained by the mass:

Work = mgh

where m is the mass (50.0 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height (4.0 m).

Plugging in the given values:

Work = (50.0 kg) x (9.8 m/s^2) x (4.0 m)
= 1960 J

The force exerted by the forklift is:

force = 1960 J / 4.0 m
= 490 N

The input energy to the forklift is given as 5200 J.

Efficiency = (useful output energy / input energy) x 100%

Efficiency = (1960 J / 5200 J) x 100%
≈ 37.7%

Therefore, the efficiency of the forklift is approximately 37.7%.