An 85-kg football player traveling is stopped in 1.0 s by a tackler. (a) What is the original kinetic energy of the player? (b) What average power is required to stop him?

The mass of the footballer is\(m = 85\;{\rm{kg}}\).

The speed of the footballer is\(v = 5\;{\rm{m/s}}\).

The time taken by the tackler to stop the player is \(t = 1\;{\rm{s}}\).

The energy possessed by a body by virtue of its motion is called kinetic energy. Mathematically, it is given by:

\(KE = \frac{1}{2}m{v^2}\) … (i)

Here, m is the mass and v is the velocity of the body.

Substitute the known numerical values in equation (i) for the kinetic energy of the player.

\(\begin{aligned}KE &= \frac{1}{2}\left( {85\;{\rm{kg}}} \right){\left( {5\;{\rm{m/s}}} \right)^2}\\ &= 1062.5\;{\rm{J}}\end{aligned}\)

Thus, the original kinetic energy of the player is 1062.5 J.

Power, P is defined as the rate at which work is done and is obtained by dividing the time taken to perform that work. It is given as:

\(P = \frac{W}{t}\)… (ii)

The average power required to stop the player is equal to the change in the energy of the player divided by the time (t = 1 s) taken to bring out the change in energy.

From equation (ii), you get:

\(\begin{aligned}P &= \frac{{1062.5\;{\rm{J}}}}{{1\;{\rm{s}}}}\\ &= 1062.5\;{\rm{W}}\end{aligned}\)

Thus, the average power is 1062.5 W.