Suggested languages for you:

Americas

Europe

Q46PE

Expert-verified
Found in: Page 262

### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# Calculate the power output in watts and horsepower of a shot-putter who takes $$1.20{\rm{ s}}$$ to accelerate the $$7.27 - {\rm{kg}}$$ shot from rest to $$14.0{\rm{ m}}/{\rm{s}}$$, while raising it $$0.800{\rm{ m}}$$. (Do not include the power produced to accelerate his body.)

The average power output of the shot-putter is $$641.26{\rm{ W}}$$ or $$0.86{\rm{ hp}}$$.

See the step by step solution

## Step 1: Definition of Concepts

The net work done on the system is given as,

$$W = \Delta KE + \Delta PE$$

Here, $$\Delta KE$$ is the change in kinetic energy, and $$\Delta PE$$ is the change in potential energy.

The power output of the system is given as,

$$P = \frac{W}{T}$$

Here, $$T$$ is the time required to do the work.

## Step 2: Calculate the power output of shot-putter in watts and horsepower

The change in kinetic energy is given as,

$$\Delta KE = \frac{1}{2}m\left( {v_f^2 - v_i^2} \right)$$

Here, $$m$$ is the mass of the shot $$\left( {m = 7.27{\rm{ kg}}} \right)$$, $${v_f}$$ is the final velocity of the shot $$\left( {{v_f} = 14.0{\rm{ m}}/{\rm{s}}} \right)$$, and $${v_i}$$ is the initial velocity of the shot ($${v_i} = 0$$ as the shot starts from rest).

Putting all known values,

$$\begin{array}{c}\Delta KE = \frac{1}{2} \times \left( {7.27{\rm{ kg}}} \right) \times \left[ {{{\left( {14.0{\rm{ m}}/{\rm{s}}} \right)}^2} - {{\left( 0 \right)}^2}} \right]\\ = 712.46{\rm{ J}}\end{array}$$

The change in potential energy is given as,

$$\Delta PE = mg\left( {{h_f} - {h_i}} \right)$$

Here, $$m$$ is the mass of the shot $$\left( {m = 7.27{\rm{ kg}}} \right)$$, $$g$$ is the acceleration due to gravity $$\left( {g = 9.81{\rm{ m}}/{{\rm{s}}^2}} \right)$$, $${h_f}$$ is the height of the shot lifted $$\left( {{h_f} = 0.800{\rm{ m}}} \right)$$, and $${h_i}$$ is the initial height of the shot $$\left( {{h_i} = 0} \right)$$.

Putting all known values,

$$\begin{array}{c}\Delta PE = \left( {7.27{\rm{ kg}}} \right) \times \left( {9.81{\rm{ m}}/{{\rm{s}}^2}} \right) \times \left[ {\left( {0.800{\rm{ m}}} \right) - \left( 0 \right)} \right]\\ = 57.05{\rm{ J}}\end{array}$$

The total work done by the shot-putter can be calculated using equation (1.1).

Putting all known values into equation (1.1),

$$\begin{array}{c}W = \left( {712.46{\rm{ J}}} \right) + \left( {57.05{\rm{ J}}} \right)\\ = 769.51{\rm{ J}}\end{array}$$

The shot-putter takes a time of $$t = 1.20{\rm{ s}}$$ to do the work.

The power output of the shot-putter can be calculated using equation (1.2).

Putting all known values into equation (1.2).

$$\begin{array}{c}P = \frac{{769.51{\rm{ J}}}}{{1.20{\rm{ s}}}}\\ = 641.26{\rm{ W}}\end{array}$$

Converting power output of the shot-putter from watts to horsepower.

$$\begin{array}{c}P = \left( {641.26{\rm{ W}}} \right) \times \left( {\frac{{1{\rm{ hp}}}}{{746{\rm{ W}}}}} \right)\\ = 0.86{\rm{ hp}}\end{array}$$

Therefore, the required average power output of the shot-putter is $$641.26{\rm{ W}}$$ or $$0.86{\rm{ hp}}$$.