A pitcher gives a power of P = 207 W to a baseball of mass m = 0.085 kg for a period of t = 0.25...

Question:

A pitcher gives a power of P = 207 W to a baseball of mass m = 0.085 kg for a period of t = 0.25 s. What is the ball's speed in m/s when it leaves his hand?

Power

In physics, power is stated as the work done produced or consumed in one unit of time. The power is calculated in watt and kilowatt. The power is also referred as the energy consumed or produced with respect to time.

Answer and Explanation:


Given data

  • The power is: {eq}P = 207\;{\rm{W}} {/eq}
  • The mass is: {eq}m = 0.085\;{\rm{kg}} {/eq}
  • The time taken is: {eq}t = 0.25\;{\rm{s}} {/eq}


The work done is,

$$W = Pt $$


The kinetic energy is,

$$KE = \dfrac{{m{V^2}}}{2} $$


The expression for the conservation of energy is,

$$KE = W $$


Substitute the known values.

$$Pt = \dfrac{{m{V^2}}}{2} $$


Substitute the known values.

$$\begin{align*} \left( {207\;{\rm{W}}} \right)\left( {0.25\;{\rm{s}}} \right) &= \dfrac{{\left( {0.085\;{\rm{kg}}} \right){V^2}}}{2}\\ {V^2} &= \dfrac{{2\left( {207\;{\rm{W}}\left( {\dfrac{{1\;{\rm{J/s}}}}{{1\;{\rm{W}}}}} \right)} \right)\left( {0.25\;{\rm{s}}} \right)}}{{\left( {0.085\;{\rm{kg}}} \right)}}\\ V &= 34.894\;{\rm{m/s}} \end{align*} $$


Thus, the speed of the ball is {eq}34.894\;{\rm{m/s}} {/eq}.


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