Here are the solutions to the questions: a) The velocity ratio (VR) of a machine is defined as the ratio of the distance moved by the effort to the distance moved by the load. $$ \text{Velocity Ratio} = \frac{\text{Distance moved by effort}}{\text{Distance moved by load}} $$ b) Given: Mass of load, $m = 20 \, \text{kg}$ Effort force, $F = 150 \, \text{N}$ Vertical height, $h = 3 \, \text{m}$ Length of inclined plane (distance moved by effort), $L = 5 \, \text{m}$ Acceleration due to gravity, $g = 10 \, \text{m/s}^2$ (assumed) i) The work done on the load Step 1: Calculate the weight of the load. The weight of the load is the force of gravity acting on it. $$ \text{Weight} = m \times g $$ $$ \text{Weight} = 20 \, \text{kg} \times 10 \, \text{m/s}^2 $$ $$ \text{Weight} = 200 \, \text{N} $$ Step 2: Calculate the work done on the load. Work done on the load is the product of its weight and the vertical height it is lifted. $$ W_{\text{load}} = \text{Weight} \times h $$ $$ W_{\text{load}} = 200 \, \text{N} \times 3 \, \text{m} $$ $$ W_{\text{load}} = 600 \, \text{J} $$ The work done on the load is $\boxed{\text{600 J}}$. ii) Work done by the effort Step 1: Identify the effort force and the distance moved by the effort. Effort force, $F = 150 \, \text{N}$ Distance moved by effort (length of the inclined plane), $L = 5 \, \text{m}$ Step 2: Calculate the work done by the effort. Work done by the effort is the product of the effort force and the distance it moves. $$ W_{\text{effort}} = F \times L $$ $$ W_{\text{effort}} = 150 \, \text{N} \times 5 \, \text{m} $$ $$ W_{\text{effort}} = 750 \, \text{J} $$ The work done by the effort is $\boxed{\text{750 J}}$.