   Chapter 5, Problem 82AP

Chapter
Section
Textbook Problem

As a 75.0-kg man steps onto a bathroom scale, the spring inside the scale compresses by 0.650 mm. Excited to see that he has lost 2.50 kg since his previous weigh-in, the man jumps 0.300 m straight up into the air and lands directly on the scale. (a) What is the spring’s maximum compression? (b) If the scale reads in kilograms, what reading does it give when the spring is at its maximum compression?

(a)

To determine
The maximum compression of the spring.

Explanation

Section1:

To determine: The spring constant.

Answer: The spring constant of the spring is 1.13×106N/m .

Explanation:

Given Info:

The mass of the man is 75.0kg .

The compression of the spring with the mass of the man is 0.650mm .

The wait lose of the man is 2.50kg .

The length that the man jump straight up is 0.300m

When the man is standing on top of the spring the total spring force acting in the upward direction will be the same as the weight of the man in the downward direction.

Fs=mg

• Fs is the spring force
• m is the mass of man
• g is acceleration due gravity

kx=mg

On re-arranging,

k=mgx

Substitute 75.0kg for m, 9.8m/s2 for g and 0.650mm for x to find the spring constant,

k=(75.0kg)(9.8m/s2)(6.5×104m)=1.13×106N/m

Thus, the spring constant of the spring is 1.13×106N/m .

Conclusion:

From section1,

The spring constant of the spring is 1.13×106N/m .

From the conservation of mechanical energy,

KEi+PEg,i+PEs,i=KEf+PEg,f+PEs,f      (I)

• KEi is the initial kinetic energy
• PEg,i is the initial potential energy due to mass of the man
• PEs,i is the initial potential due to spring
• KEf is the final kinetic energy
• PEg,f is the initial potential energy due to mass of the man
• PEs,f is the final potential energy due to spring

Consider the equilibrium point as h=0 point

(b)

To determine
The reading in kilogram that the scale gives when the spring is at its maximum compression.

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