Physics for Scientists and Engineers with Modern, Revised Hybrid (with Enhanced WebAssign Printed Access Card for Physics, Multi-Term Courses)
Physics for Scientists and Engineers with Modern, Revised Hybrid (with Enhanced WebAssign Printed Access Card for Physics, Multi-Term Courses)
9th Edition
ISBN: 9781305266292
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 5, Problem 41P

(a)

To determine

The force exerted by the chin-up bar on the person’s body at t=0.

(a)

Expert Solution
Check Mark

Answer to Problem 41P

The force exerted by the chin-up bar on the person’s body at t=0 is 646 N up

Explanation of Solution

The slope of the speed versus time graph gives the magnitude of the acceleration. Consider positive slope of speed versus time graph as positive acceleration and negative slope of speed versus time graph gives negative acceleration. When the slope is zero the acceleration is zero.

According to Newton’s second law,

  may=Fbarmg

Here, Fbar is the force exerted by the chin-up bar on the body, m is the mass of the person, g is the acceleration due to gravity, and ay is the vertical acceleration.

Rewrite the above equation in terms of Fbar

  Fbar=mg+may=m(g+ay)                                                                                                  (I)

Write the expression to calculate the acceleration.

    ay=sfsitfti                                                                                                    (II)

Here, si is the initial speed, sf is the final speed, ti is the initial time, and tf is the final time.

Physics for Scientists and Engineers with Modern, Revised Hybrid (with Enhanced WebAssign Printed Access Card for Physics, Multi-Term Courses), Chapter 5, Problem 41P , additional homework tip  1

From the graph the slope at time t=0.

sf=s2, si=s1, tf=t2, and ti=t1

Conclusion:

Substitute 6cm/s for si, 18cm/s for sf, 0.2s for ti, 0.6s for tf in (II) to find ay.

    ay=18cm/s6cm/s0.6s0.2s=12cm/s0.4s=30cm/s2(102m1cm)=0.3 m/s2 s

Substitute 0.3 m/s2 for ay , 9.80 m/s2 for g , and 64.0kg for m in (I) to find Fbar.

    Fbar=(64.0 kg)(9.80 m/s2+0.3 m/s2)=646.4 N up646 N up

Therefore, the force exerted by the chin-up bar on the person’s body at t=0 is 646 N up

(b)

To determine

The force exerted by the chin-up bar on the person’s body at t=0.5s.

(b)

Expert Solution
Check Mark

Answer to Problem 41P

The force exerted by the chin-up bar on the person’s body at t=0.5s is 646 N up

Explanation of Solution

The slope of the speed versus time graph gives the magnitude of the acceleration. The slope of the graph at 0.5 s is same as that of 0 s. That is the graph is a straight line at these moment. If the graph is a straight line, the acceleration is same.

The mass of the person does not change and the acceleration is same as that at t=0s

Conclusion:

Therefore, the force exerted by the chin-up bar on the person’s body at t=0.5s is 646 N up

(c)

To determine

The force exerted by the chin-up bar on the person’s body at t=1.1s.

(c)

Expert Solution
Check Mark

Answer to Problem 41P

The force exerted by the chin-up bar on the person’s body at t=1.1s is 627Nup

Explanation of Solution

The graph below is the speed versus time graph

Physics for Scientists and Engineers with Modern, Revised Hybrid (with Enhanced WebAssign Printed Access Card for Physics, Multi-Term Courses), Chapter 5, Problem 41P , additional homework tip  2

From the graph it is evident that at t=1.1s, the slope is zero. Thus, the acceleration is zero.

Conclusion:

Substitute 0 m/s2 for ay , 9.80 m/s2 for g , and 64.0kg for m in (I) to find Fbar.

    Fbar=(64.0 kg)(9.80 m/s2+0 m/s2)=627.2 Nup627 Nup

Therefore, the force exerted by the chin-up bar on the person’s body at t=1.1s is 627 Nup

(d)

To determine

The force exerted by the chin-up bar on the person’s body at t=1.6s.

(d)

Expert Solution
Check Mark

Answer to Problem 41P

The force exerted by the chin-up bar on the person’s body at t=1.6s is 593 N up

Explanation of Solution

The graph below is the speed versus time graph.

Physics for Scientists and Engineers with Modern, Revised Hybrid (with Enhanced WebAssign Printed Access Card for Physics, Multi-Term Courses), Chapter 5, Problem 41P , additional homework tip  3

From the graph the slope at time t=1.6s.

sf=s4, si=s3, tf=t4, and ti=t3

Conclusion:

Substitute 11cm/s for si, 2cm/s for sf, 1.5s for ti, 1.67s for tf in (II) to find ay.

    ay=2cm/s11cm/s1.67s1.5s=9cm/s0.17s=53cm/s2(102m1cm)=0.53 m/s2

Substitute 0.53 m/s2 for ay , 9.80 m/s2 for g , and 64.0kg for m in (I) to find Fbar.

    Fbar=(64.0 kg)(9.80 m/s20.53 m/s2)=593 N up

Therefore, the force exerted by the chin-up bar on the person’s body at t=1.6s is 593 N up

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Chapter 5 Solutions

Physics for Scientists and Engineers with Modern, Revised Hybrid (with Enhanced WebAssign Printed Access Card for Physics, Multi-Term Courses)

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