Question

During the first step of a sprint race a sprinter of mass 64kg generates a horizontal braking impulse of -3.2 Ns and a horizontal propulsive impulse of +56.0 Ns.
What is the net horizontal mechanical work done by the sprinter in the first step?

Answer 1

During the first step of a sprint race a sprinter of mass 64kg generates a horizontal braking impulse of -3.2 Ns and a horizontal propulsive impulse of +56.0 Ns. What is the net horizontal mechanical work done by the sprinter in the first step?

**Solution.**

$$m = 64kg, p_1 = 56.0\ \text{Ns}, p_2 = -3.2\ \text{Ns};$$ $$W - ?$$

The mechanical work is equal to the change in kinetic energy of the sprinter:

$$W = \Delta E_k.$$ $$\Delta E_k = E_{kf} - E_{ki};$$

$E_{ki}$ - the initial kinetic energy of the sprinter;

$E_{kf}$ - the final kinetic energy of the sprinter.

A sprinter is at rest then the initial kinetic energy is:

$$E_{ki} = 0.$$

The final kinetic energy of the sprinter is:

$$E_{kf} = \frac{m v_f^2}{2}.$$

Multiply the numerator and denominator by $m$:

$$E_{kf} = \frac{m v_f^2 m}{2m};$$ $$E_{kf} = \frac{m^2 v_f^2}{2m}.$$

The impulse of the object is $p = mv$, then the final kinetic energy of the sprinter is:

$$E_{kf} = \frac{p_f^2}{2m}.$$

The initial impulse of the sprinter is:

$$p_i = 0.$$

The final impulse of the sprinter is:

$$p_f = p_1 + p_2.$$

The final kinetic energy of the sprinter is:

$$E _ {k f} = \frac {(p _ {1} + p _ {2}) ^ {2}}{2 m}.$$

The mechanical work is:

$$W = \frac {(p _ {1} + p _ {2}) ^ {2}}{2 m}.$$ $$W = \frac {\left(56Ns + (-3.2Ns)\right) ^ {2}}{2 \cdot 64kg} = \frac {(56Ns - 3.2Ns) ^ {2}}{2 \cdot 64kg} = 21.78J.$$

**Answer:**

The net horizontal mechanical work done by the sprinter in the first step is $W = 21.78J$.