Question

500 J of work is required by a machine to raise a load of 250 N through a vertical distance of 1.6 m. What is the efficiency of the machine?

Answer 1

Let's first calculate the work done in raising a load:

$$W=Fd=250\ N\cdot1.6\ m=400\ J.$$

Finally we can find the efficiency of the machine:

$$\eta=\dfrac{useful\ output\ energy}{input\ energy}\cdot100\%,$$ $$\eta=\dfrac{400\ J}{500\ J}\cdot100\%=80\%.$$

Answer:

$\eta=80\%.$