Calculate acceleration (a) and final velocity (v_f).

Question

Calculate acceleration (a) and final velocity (v_f).

Asked on March 29, 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 · 2026-03-29T15:07:07.152Z

13. An object of mass 24 kg is accelerated up a frictionless plane inclined at an angle of 37° with the horizontal by a constant force. Starting at the bottom from the rest, it covers a distance of 18m in 3.0secs. Given: $m = 24 \text{ kg}$, $\theta = 37^\circ$, $v_0 = 0 \text{ m/s}$, $d = 18 \text{ m}$, $t = 3.0 \text{ s}$. Assume $g = 9.8 \text{ m/s}^2$. Step 1: Calculate acceleration ($a$) and final velocity ($v_f$). $$ d = v_0 t + \frac{1}{2} a t^2 \implies 18 = 0 \cdot 3 + \frac{1}{2} a (3)^2 $$ $$ 18 = 4.5a \implies a = \frac{18}{4.5} = 4.0 \text{ m/s}^2 $$ $$ v_f = v_0 + at \implies v_f = 0 + (4.0)(3.0) = 12 \text{ m/s} $$ Step 2: Calculate the applied force ($F_{app}$). The net force along the incline is $F_{net} = ma$. The forces are the applied force and the component of gravity along the incline ($mg\sin\theta$). $$ F_{app} - mg\sin\theta = ma $$ $$ F_{app} = ma + mg\sin\theta = (24 \text{ kg})(4.0 \text{ m/s}^2) + (24 \text{ kg})(9.8 \text{ m/s}^2)\sin(37^\circ) $$ $$ F_{app} = 96 \text{ N} + 235.2 \text{ N} \cdot 0.6018 \approx 96 \text{ N} + 141.56 \text{ N} = 237.56 \text{ N} $$ a) What is the average power required to accomplish the process? Step 3: Calculate the work done ($W$) and average power ($P_{avg}$). $$ W = F_{app} \cdot d = (237.56 \text{ N})(18 \text{ m}) = 4276.08 \text{ J} $$ $$ P_{avg} = \frac{W}{t} = \frac{4276.08 \text{ J}}{3.0 \text{ s}} = 1425.36 \text{ W} $$ The average power is $\boxed{1.43 \times 10^3 \text{ W}}$. b) What is the instantaneous power required at the end of the 3.0 second interval? Step 4: Calculate the instantaneous power ($P_{inst}$). $$ P_{inst} = F_{app} \cdot v_f = (237.56 \text{ N})(12 \text{ m/s}) = 2850.72 \text{ W} $$ The instantaneous power is $\boxed{2.85 \times 10^3 \text{ W}}$. 14. A body is pulled along a horizontal force of 20N, calculate the work done in moving a distance of 40m. Given: $F = 20 \text{ N}$, $d = 40 \text{ m}$. Step 1: Calculate the work done ($W$). $$ W = F \cdot d $$ $$ W = (20 \text{ N})(40 \text{ m}) = 800 \text{ J} $$ The work done is $\boxed{800 \text{ J}}$. 15. What is the velocity ratio of an inclined plane of length 6m if the higher end of the plane is 2m above the ground? Given: Length of inclined plane $L = 6 \text{ m}$, Height $h = 2 \text{ m}$. Step 1: Calculate the velocity ratio (VR). $$ VR = \frac{\text{Length of incline}}{\text{Vertical height}} $$ $$ VR = \frac{6 \text{ m}}{2 \text{ m}} = 3 $$ The velocity ratio is $\boxed{3}$. 16. A machine of efficiency of 70% is used to raise a body of mass 80kg through a vertical distance of 3m in 40 seconds. Calculate the power input. (Take g = 10 m/s²) Given: $\eta = 70\% = 0.70$, $m = 80 \text{ kg}$, $h = 3 \text{ m}$, $t = 40 \text{ s}$, $g = 10 \text{ m/s}^2$. Step 1: Calculate the work output ($W_{out}$). $$ W_{out} = mgh = (80 \text{ kg})(10 \text{ m/s}^2)(3 \text{ m}) = 2400 \text{ J} $$ Step 2: Calculate the power output ($P_{out}$). $$ P_{out} = \frac{W_{out}}{t} = \frac{2400 \text{ J}}{40 \text{ s}} = 60 \text{ W} $$ Step 3: Calculate the power input ($P_{in}$). $$ \eta = \frac{P_{out}}{P_{in}} \implies P_{in} = \frac{P_{out}}{\eta} $$ $$ P_{in} = \frac{60 \text{ W}}{0.70} \approx 85.71 \text{ W} $$ The power input is $\boxed{85.7 \text{ W}}$. 17. (i) Give an example of a machine where the load is between the fulcrum and the effort. (ii) A wheel and axle is used to raise a man whose weight is 800N by the application of 250N. The radii of the wheel and axle are 800mm and 200mm respectively, what is the efficiency of the machine? i) Example of a machine where the load is between the fulcrum and the effort. This describes a Class 2 lever. Example: A wheelbarrow. ii) Efficiency of a wheel and axle. Given: Load $F_L = 800 \text{ N}$, Effort $F_E = 250 \text{ N}$, Wheel radius $R_W = 800 \text{ mm} = 0.8 \text{ m}$, Axle radius $R_A = 200 \text{ mm} = 0.2 \text{ m}$. Step 1: Calculate the Mechanical Advantage (MA). $$ MA = \frac{F_L}{F_E} = \frac{800 \text{ N}}{250 \text{ N}} = 3.2 $$ Step 2: Calculate the Velocity Ratio (VR). $$ VR = \frac{R_W}{R_A} = \frac{0.8 \text{ m}}{0.2 \text{ m}} = 4 $$ Step 3: Calculate the efficiency ($\eta$). $$ \eta = \frac{MA}{VR} \times 100\% $$ $$ \eta = \frac{3.2}{4} \times 100\% = 0.8 \times 100\% = 80\% $$ The efficiency of the machine is $\boxed{80\%}$. 18. A wheel and axle have radii 80cm and 10 cm respectively. If the efficiency of the machine is 0.85, what load will an applied force of 1200N to the wheel raise? Given: Wheel radius $R_W = 80 \text{ cm} = 0.8 \text{ m}$, Axle radius $R_A =

Accepted Answer · 2026-03-29T15:07:07.152Z

13. An object of mass 24 kg is accelerated up a frictionless plane inclined at an angle of 37° with the horizontal by a constant force. Starting at the bottom from the rest, it covers a distance of 18m in 3.0secs. Given: $m = 24 \text{ kg}$, $\theta = 37^\circ$, $v_0 = 0 \text{ m/s}$, $d = 18 \text{ m}$, $t = 3.0 \text{ s}$. Assume $g = 9.8 \text{ m/s}^2$. Step 1: Calculate acceleration ($a$) and final velocity ($v_f$). $$ d = v_0 t + \frac{1}{2} a t^2 \implies 18 = 0 \cdot 3 + \frac{1}{2} a (3)^2 $$ $$ 18 = 4.5a \implies a = \frac{18}{4.5} = 4.0 \text{ m/s}^2 $$ $$ v_f = v_0 + at \implies v_f = 0 + (4.0)(3.0) = 12 \text{ m/s} $$ Step 2: Calculate the applied force ($F_{app}$). The net force along the incline is $F_{net} = ma$. The forces are the applied force and the component of gravity along the incline ($mg\sin\theta$). $$ F_{app} - mg\sin\theta = ma $$ $$ F_{app} = ma + mg\sin\theta = (24 \text{ kg})(4.0 \text{ m/s}^2) + (24 \text{ kg})(9.8 \text{ m/s}^2)\sin(37^\circ) $$ $$ F_{app} = 96 \text{ N} + 235.2 \text{ N} \cdot 0.6018 \approx 96 \text{ N} + 141.56 \text{ N} = 237.56 \text{ N} $$ a) What is the average power required to accomplish the process? Step 3: Calculate the work done ($W$) and average power ($P_{avg}$). $$ W = F_{app} \cdot d = (237.56 \text{ N})(18 \text{ m}) = 4276.08 \text{ J} $$ $$ P_{avg} = \frac{W}{t} = \frac{4276.08 \text{ J}}{3.0 \text{ s}} = 1425.36 \text{ W} $$ The average power is $\boxed{1.43 \times 10^3 \text{ W}}$. b) What is the instantaneous power required at the end of the 3.0 second interval? Step 4: Calculate the instantaneous power ($P_{inst}$). $$ P_{inst} = F_{app} \cdot v_f = (237.56 \text{ N})(12 \text{ m/s}) = 2850.72 \text{ W} $$ The instantaneous power is $\boxed{2.85 \times 10^3 \text{ W}}$. 14. A body is pulled along a horizontal force of 20N, calculate the work done in moving a distance of 40m. Given: $F = 20 \text{ N}$, $d = 40 \text{ m}$. Step 1: Calculate the work done ($W$). $$ W = F \cdot d $$ $$ W = (20 \text{ N})(40 \text{ m}) = 800 \text{ J} $$ The work done is $\boxed{800 \text{ J}}$. 15. What is the velocity ratio of an inclined plane of length 6m if the higher end of the plane is 2m above the ground? Given: Length of inclined plane $L = 6 \text{ m}$, Height $h = 2 \text{ m}$. Step 1: Calculate the velocity ratio (VR). $$ VR = \frac{\text{Length of incline}}{\text{Vertical height}} $$ $$ VR = \frac{6 \text{ m}}{2 \text{ m}} = 3 $$ The velocity ratio is $\boxed{3}$. 16. A machine of efficiency of 70% is used to raise a body of mass 80kg through a vertical distance of 3m in 40 seconds. Calculate the power input. (Take g = 10 m/s²) Given: $\eta = 70\% = 0.70$, $m = 80 \text{ kg}$, $h = 3 \text{ m}$, $t = 40 \text{ s}$, $g = 10 \text{ m/s}^2$. Step 1: Calculate the work output ($W_{out}$). $$ W_{out} = mgh = (80 \text{ kg})(10 \text{ m/s}^2)(3 \text{ m}) = 2400 \text{ J} $$ Step 2: Calculate the power output ($P_{out}$). $$ P_{out} = \frac{W_{out}}{t} = \frac{2400 \text{ J}}{40 \text{ s}} = 60 \text{ W} $$ Step 3: Calculate the power input ($P_{in}$). $$ \eta = \frac{P_{out}}{P_{in}} \implies P_{in} = \frac{P_{out}}{\eta} $$ $$ P_{in} = \frac{60 \text{ W}}{0.70} \approx 85.71 \text{ W} $$ The power input is $\boxed{85.7 \text{ W}}$. 17. (i) Give an example of a machine where the load is between the fulcrum and the effort. (ii) A wheel and axle is used to raise a man whose weight is 800N by the application of 250N. The radii of the wheel and axle are 800mm and 200mm respectively, what is the efficiency of the machine? i) Example of a machine where the load is between the fulcrum and the effort. This describes a Class 2 lever. Example: A wheelbarrow. ii) Efficiency of a wheel and axle. Given: Load $F_L = 800 \text{ N}$, Effort $F_E = 250 \text{ N}$, Wheel radius $R_W = 800 \text{ mm} = 0.8 \text{ m}$, Axle radius $R_A = 200 \text{ mm} = 0.2 \text{ m}$. Step 1: Calculate the Mechanical Advantage (MA). $$ MA = \frac{F_L}{F_E} = \frac{800 \text{ N}}{250 \text{ N}} = 3.2 $$ Step 2: Calculate the Velocity Ratio (VR). $$ VR = \frac{R_W}{R_A} = \frac{0.8 \text{ m}}{0.2 \text{ m}} = 4 $$ Step 3: Calculate the efficiency ($\eta$). $$ \eta = \frac{MA}{VR} \times 100\% $$ $$ \eta = \frac{3.2}{4} \times 100\% = 0.8 \times 100\% = 80\% $$ The efficiency of the machine is $\boxed{80\%}$. 18. A wheel and axle have radii 80cm and 10 cm respectively. If the efficiency of the machine is 0.85, what load will an applied force of 1200N to the wheel raise? Given: Wheel radius $R_W = 80 \text{ cm} = 0.8 \text{ m}$, Axle radius $R_A =