Determine the velocity ratio of the machine.

Question

Determine the velocity ratio of the machine.

Asked on April 24, 2026Physics

This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.

Accepted Answer · 2026-04-24T09:34:33.267Z

Step 1: Determine the velocity ratio of the machine. Based on the pulley system shown in the previous question, there are 6 rope segments supporting the movable block. Therefore, the velocity ratio (VR) is: $$\text{VR} = \boxed{6}$$ Step 2: Determine the efficiency of the pulley system. Given: Load, $L = 500 \, \text{N}$ (from previous question) Effort, $E = 150 \, \text{N}$ Velocity Ratio, VR = 6 First, calculate the Mechanical Advantage (MA): $$\text{MA} = \frac{\text{Load}}{\text{Effort}}$$ $$\text{MA} = \frac{500 \, \text{N}}{150 \, \text{N}}$$ $$\text{MA} = \frac{10}{3} \approx 3.333$$ Next, calculate the efficiency ($\eta$): $$\eta = \frac{\text{MA}}{\text{VR}} \times 100\%$$ $$\eta = \frac{10/3}{6} \times 100\%$$ $$\eta = \frac{10}{18} \times 100\%$$ $$\eta = \frac{5}{9} \times 100\%$$ $$\eta \approx 55.55\%$$ Rounding to three significant figures: $$\eta = \boxed{55.6\%}$$ Step 3: Sketch a graph of efficiency against load for the system. The provided sketch shows a typical graph of efficiency versus load for a machine. The efficiency generally increases with the load up to a certain point, after which it tends to level off. This is because the constant frictional forces become a smaller proportion of the total work done as the load increases. Step 4: Determine how high the block will rise. Given: Force, $F = 20 \, \text{N}$ Mass, $m = 1 \, \text{kg}$ Time, $t = 1.2 \, \text{s}$ Acceleration due to gravity, $g = 9.8 \, \text{m/s}^2$ First, calculate the acceleration $a$ of the block using Newton's second law: $$F = ma$$ $$a = \frac{F}{m}$$ $$a = \frac{20 \, \text{N}}{1 \, \text{kg}}$$ $$a = 20 \, \text{m/s}^2$$ Next, calculate the velocity $v$ of the block after the force acts for $1.2 \, \text{s}$ (assuming it starts from rest): $$v = u + at$$ $$v = 0 + (20 \, \text{m/s}^2)(1.2 \, \text{s})$$ $$v = 24 \, \text{m/s}$$ Finally, use the conservation of energy to find the maximum height $h$ the block rises. The kinetic energy at the bottom of the swing is converted into potential energy at the maximum height: $$\frac{1}{2}mv^2 = mgh$$ Divide both sides by $m$: $$\frac{1}{2}v^2 = gh$$ Solve for $h$: $$h = \frac{v^2}{2g}$$ $$h = \frac{(24 \, \text{m/s})^2}{2 \times 9.8 \, \text{m/s}^2}$$ $$h = \frac{576 \, \text{m}^2/\text{s}^2}{19.6 \, \text{m/s}^2}$$ $$h \approx 29.3877 \, \text{m}$$ Rounding to three significant figures: $$h = \boxed{29.4 \, \text{m}}$$ 3 done, 2 left today. You're making progress.

Accepted Answer · 2026-04-24T09:34:33.267Z

Step 1: Determine the velocity ratio of the machine. Based on the pulley system shown in the previous question, there are 6 rope segments supporting the movable block. Therefore, the velocity ratio (VR) is: $$\text{VR} = \boxed{6}$$ Step 2: Determine the efficiency of the pulley system. Given: Load, $L = 500 \, \text{N}$ (from previous question) Effort, $E = 150 \, \text{N}$ Velocity Ratio, VR = 6 First, calculate the Mechanical Advantage (MA): $$\text{MA} = \frac{\text{Load}}{\text{Effort}}$$ $$\text{MA} = \frac{500 \, \text{N}}{150 \, \text{N}}$$ $$\text{MA} = \frac{10}{3} \approx 3.333$$ Next, calculate the efficiency ($\eta$): $$\eta = \frac{\text{MA}}{\text{VR}} \times 100\%$$ $$\eta = \frac{10/3}{6} \times 100\%$$ $$\eta = \frac{10}{18} \times 100\%$$ $$\eta = \frac{5}{9} \times 100\%$$ $$\eta \approx 55.55\%$$ Rounding to three significant figures: $$\eta = \boxed{55.6\%}$$ Step 3: Sketch a graph of efficiency against load for the system. The provided sketch shows a typical graph of efficiency versus load for a machine. The efficiency generally increases with the load up to a certain point, after which it tends to level off. This is because the constant frictional forces become a smaller proportion of the total work done as the load increases. Step 4: Determine how high the block will rise. Given: Force, $F = 20 \, \text{N}$ Mass, $m = 1 \, \text{kg}$ Time, $t = 1.2 \, \text{s}$ Acceleration due to gravity, $g = 9.8 \, \text{m/s}^2$ First, calculate the acceleration $a$ of the block using Newton's second law: $$F = ma$$ $$a = \frac{F}{m}$$ $$a = \frac{20 \, \text{N}}{1 \, \text{kg}}$$ $$a = 20 \, \text{m/s}^2$$ Next, calculate the velocity $v$ of the block after the force acts for $1.2 \, \text{s}$ (assuming it starts from rest): $$v = u + at$$ $$v = 0 + (20 \, \text{m/s}^2)(1.2 \, \text{s})$$ $$v = 24 \, \text{m/s}$$ Finally, use the conservation of energy to find the maximum height $h$ the block rises. The kinetic energy at the bottom of the swing is converted into potential energy at the maximum height: $$\frac{1}{2}mv^2 = mgh$$ Divide both sides by $m$: $$\frac{1}{2}v^2 = gh$$ Solve for $h$: $$h = \frac{v^2}{2g}$$ $$h = \frac{(24 \, \text{m/s})^2}{2 \times 9.8 \, \text{m/s}^2}$$ $$h = \frac{576 \, \text{m}^2/\text{s}^2}{19.6 \, \text{m/s}^2}$$ $$h \approx 29.3877 \, \text{m}$$ Rounding to three significant figures: $$h = \boxed{29.4 \, \text{m}}$$ 3 done, 2 left today. You're making progress.

Determine the velocity ratio of the machine.