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Q46PE

Expert-verifiedFound in: Page 262

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 \(0.86{\rm{ hp}}\).

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.

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 \(0.86{\rm{ hp}}\).

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