Q47P - University Physics with Modern Physics

Question

A 75.0-kg man steps off a platform 3.10 m above the ground. He keeps his legs straight as he falls, but his knees begin to bend at the moment his feet touch the ground; treated as a particle, he moves an additional  0.60 m before coming to rest. (a) What is his speed at the instant his feet touch the ground? (b) If we treat the man as a particle, what is his acceleration (magnitude and direction) as he slows down, if the acceleration is assumed to be constant? (c) Draw his free body diagram. In terms of the forces on the diagram, what is the net force on him? Use Newton’s laws and the results of part (b) to calculate the average force his feet exert on the ground while he slows down. Express this force both in newtons and as a multiple of his weight.

Step-by-Step Solution

Answer

(a) His speed at the instant is 7.79 m/s

(b) His acceleration is $-50.0\mathrm{m}/{\mathrm{s}}^2$ and it is the upward acceleration.

(c) .

The net force on him is W-N.

The average force his feet exert in Newton is 4522.5 N.

The force as a multiple of his weight is 6.15mg .

Step 1: Identification of the given data

The given data is listed below as:

 

• The mass of the man is, m=75.0 kg
• The distance of the platform above the ground is, s=3.10 m
• The additional distance moved by the person is, $s_1=0.60 m$

Step 2: Significance of the force

The force exerted by an object is equal to the product of the mass and the acceleration of that object. The force is also proportional to the acceleration due to gravity.

Step 3: (a) Determination of the speed at the instant

The equation of the speed at the instant is expressed as:

 

$v^2=u^2+2gs$ 

 

Here, v is the final and u is the initial speed of the man. g is the acceleration due to gravity and s is the distance of the platform above the ground.

 

As initially the man was at rest, hence the initial speed of the man is zero.

 

Substitute the values in the above equation.

$$\begin{gathered} v^2=0+2\left(9.8 m/s^2\right)\left(3.10m\right) \\ =\left(19.6 m/s^2\right)\left(3.10 m\right) \\ =60.76 m^2/s^2 \\ v=7.79 m/s \end{gathered}$$ 

 

The negative value is neglected as the velocity cannot be negative.

 

Thus, his speed at the instant is 7.79 m/s.

Step 4: (b) Determination of the acceleration

The equation of the acceleration is expressed as:

 

$v^2=u^2+2a{s}_1$ 

 

Here, v is the final and u is the initial speed of the man. a is the acceleration of the many and $s_1$ is the additional distance moved by the person.

 

As finally, the man came to rest, hence the final speed of the man is zero.

 

Substitute the values in the above equation.

 

$$\begin{gathered} 0={\left(7.79 \mathrm{m}/\mathrm{s}\right)}^2+2\mathrm{a}\left(0.60 \mathrm{m}\right) \\ \left(1.2 \mathrm{m}\right)\mathrm{a}=-60.68 {\mathrm{m}}^2/{\mathrm{s}}^2 \\ \mathrm{a}=-50.5 \mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$ 

 

As the acceleration is negative, then the acceleration is in the upward acceleration.

 

Thus, his acceleration is $-50.5 \mathrm{m}/{\mathrm{s}}^2$ and it is the upward acceleration.

Step 5: (c) Determination of the free body diagram

The free body diagram of the man is drawn below:

Here, in the above diagram, the weight of the man W is acting downwards. The acceleration $a_y$ is acting in the upwards direction and N is the normal force acting on the man.

Step 6: (c) Determination of the net force

According to the diagram, the equation of the net force can be expressed as:

 

F =W-N

 

Here, W is weight of the man and N is the normal force acting on the man.

 

Thus, the net force on him is W-N.

Step 7: (c) Determination of the average force

The equation of the average force is expressed as:

 

mg-F=ma

 

Here, m is the mass of the man, g is the acceleration due to gravity, F is the average force and a is the acceleration of the man.

 

Substitute the values in the above equation.

$$\begin{gathered} \left(75.0 \mathrm{kg}\right)\left(9.8 \mathrm{m}/{\mathrm{s}}^2\right)-\mathrm{F}=\left(75.0 \mathrm{kg}\right)\left(-50.5 \mathrm{m}/{\mathrm{s}}^2\right) \\ 735 \mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2-\mathrm{F}=-3787.5 \mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2 \\ \mathrm{F}=735 \mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2+3787.5 \mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2 \\ =4522.5 \mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$ 

 

Hence, further as:

$$\begin{gathered} \mathrm{F}=4522.5\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2\times \left(\frac{1\mathrm{N}}{1 \mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2}\right) \\ =4522.5 \mathrm{N} \end{gathered}$$ 

 

Thus, the average force his feet exert in Newton is 4522.5 N.

Step 8: (c) Determination of the force as a multiple of his weight

The equation of the force as a multiple of his weight is expressed as:

 

$N=\frac{4522.5}{mg}$

 

Substitute all the values in the above equation.

 

$$\begin{gathered} N=\frac{4522.5}{\left(75\right)\left(9.8\right)} \\ =\frac{4522.5}{\left(735\right)} \\ =6.15 mg \end{gathered}$$

 

Thus, the force as a multiple of his weight is 6.15 mg.