1. Acrobat Art of mass m stands on the left end of a seesaw. Acrobat Bart of mass M jumps from a
height h onto the right end of the seesaw, thus propelling Art into the air.

Question

a. Neglecting inefficiencies, how will the PE of Art at the top of his trajectory compare with the PE of Bart just before Bart jumps?

b. Neglecting inefficiencies, show that Art reaches a height [latex]\frac{M}{m} h[/latex].

c. If Art’s mass is 40 kg, Bart’s mass is 70 kg, and the height of the initial jump was 4 m, show that Art rises a vertical distance of 7 m.

2. A loaded elevator is lifted a distance h in a time t by a motor delivering power P.

a. Show that the force exerted by the motor is [latex]\frac{pt}{h}[/latex].

b. If the elevator is raised 20 m in a time of 30 s, and the power of the motor is 60 kW, show tha the force exerted by the motor is 90 kN.

Accepted Answer

1. a. Neglecting inefficiencies, the entire initial PE of acrobat Bart before he drops goes into the PE of acrobat Art rising to his peak—that is, at Art’s moment of zero KE.

b. [latex]\begin{aligned}& ext { From } PE _{ Bart }= PE _{ Art } \Rightarrow M g h_{ Bart }= \&m g h_{ Art } \Rightarrow h_{A r t}=\frac{M}{m} h\end{aligned}[/latex].

c. [latex]h_{ ext {Art }}=\frac{M}{m} h=\left(\frac{70 kg }{40 kg } ight) 4 m =7 m[/latex].

2. a. [latex] ext { From Power }=\frac{ ext { Work }}{ ext { time }}=\frac{ ext { force } imes ext { distance }}{ ext { time }} ext {, }[/latex], we see that [latex]P=\frac{F h}{t}[/latex]. Rearranging, [latex]F=\frac{P t}{h}[/latex].

[latex]\begin{aligned}& ext { b. } F=\frac{P t}{h}=\frac{\left(60 imes 10^{3} W ight)(30 s )}{20 m }= \&9.0 imes 10^{4} \frac{\left(\frac{ N \cdot m }{ s } ight) s }{ m }=9.0 imes 10^{4} N =90 kN\end{aligned}[/latex]

Question 5.PS.2: 1. Acrobat Art of mass m stands on the left end of a seesaw....

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