Question

A runner whose mass is 50 kgaccelerates from a stop to a speed of10 m / s in 3 s. (A good sprinter can run100 m in about 10 s, with an average speed of 10 m / s.) (a) What is the average horizontal component of the force that the ground exerts on the runner’s shoes? (b) How much displacement is there of the force that acts on the sole of the runner’s shoes, assuming that there is no slipping? Therefore, how much work is done on the extended system (the runner) by the force you calculated in the previous exercise? How much work is done on the point particle system by this force? (c) The kinetic energy of the runner increases—what kind of energy decreases? By how much?

Short answer

a) 167N

b) There is no work done on the extended system due to zero displacement and , the work done on the particle system is 2500 J.

c) Decrease in the internal energy by 2500 J.

Step-by-step solution

Identification of the given data

The given data can be listed below as,

• The mass of the runner is, 50 kg.
• The velocity of the runner is, 10 m / s.
• The time taken by the runner is, 3 s.

Significance of the law of conservation of work done and Newton’s second law for the runner

The law of conservation of work done states that energy cannot be destroyed or created, it can only be converted into different forms.

The second law of Newton states that the force exerted is directly proportional to the product of the mass and the acceleration of a body.

According to the work-energy theorem, the work done is the same as that of the change in the kinetic energy.

Determination of force’s horizontal component exerted by the ground on the shoes of the runner (a)

The expression for the horizontal component of the force can be expressed as,

$$\begin{gathered} F=ma \\ =m\left(\frac{v}{t}\right) \end{gathered}$$

Here, m is the mass of the runner, a is the acceleration of the runner that is $a=\frac{v}{t}$ (here, v is the velocity of the runner and is the time taken by the runner).

For $m=50\mathrm{kg},v=10\mathrm{m}/\mathrm{s}$, and t = 3 s.

$$\begin{gathered} F=\left(50\mathrm{kg}\right)\times \left(\frac{10\mathrm{m}/\mathrm{s}}{3\mathrm{s}}\right) \\ =166.66\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2 \\ =166.66\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2\times \frac{1\mathrm{N}}{1\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2} \\ \approx 167\mathrm{N} \end{gathered}$$

Thus, the average horizontal component of the force is 167 N.

Determination of the displacement and the work done (b)

The expression for the displacement of the runner is expressed as,

s = ut

Here, u is the initial velocity of the runner’s shoes that is 0 and tis the time taken.

Substitute all the values in the above equation.

$$\begin{gathered} \mathrm{s}=0\mathrm{m}/\mathrm{s}\times 3\mathrm{s} \\ =0\mathrm{m} \end{gathered}$$

Thus, the displacement of the runner’s shoes is 0.

The expression for work done on the extended system can be expressed as,

$W=F.d$

Here, F is the force and is the displacement of the runner.

For F = 167 N and d = 0.

$$\begin{gathered} W=167N\times 0 \\ =0 \end{gathered}$$

Thus, the work done on the extended system is 0.

From the law of conservation of work done, the equation of the work done on the point particle system can be expressed as,

$W_1=\frac{1}{2}m{v}^2$

Here, $W_1$is the work done on the point particle system.

For m = 50 kg andv = 10 m/s.|
$$\begin{gathered} W=\frac{1}{2}\times 50\mathrm{kg}\times {(10\mathrm{m}/\mathrm{s})}^2=2500\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2 \\ =2500\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2\times \frac{1\mathrm{J}}{1\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2} \\ =2500\mathrm{J} \end{gathered}$$

Thus, the work done on the point particle system by this force is 2500 J.

Determination of the energy that decreases (c)

By the work energy theorem, the work done is same as that of the change in the kinetic energy.

As the Kinetic energy of the runner increases, the internal or the chemical energy of the runner decreases.

As the work done on the point particle system is 2500 J so the internal energy also decreased by 2500 J .

Thus, the internal energy decreases and it reduces to 2500 J .