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lana [24]
3 years ago
6

It took David 2 seconds to lift a 5 Newton bag of toys from the floor to the top of the top shelf, which is 3 meters tall. How m

uch power did David use?
A. 3.3 W

B. 6.1 W

C. 7.5 W

D. 14 W
Physics
1 answer:
just olya [345]3 years ago
7 0
So power is equal to work over time and work is force times distance, you do 5 times 3 and get 15 dividing by 2 gives us 7.5 W answer c
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I think the situation is modeled by the scenario in the attached image. Some specific values seem to be missing (like the height of door d)...

The door forms a right triangles that satisfies

\tan\theta=\dfrac ab\implies\sec^2\theta\dfrac{\mathrm d\theta}{\mathrm dt}=\dfrac{b\frac{\mathrm da}{\mathrm dt}-a\frac{\mathrm db}{\mathrm dt}}{b^2}

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so if you happen to know the height of the door, you can solve for b and a.

d is fixed, so

a^2+b^2=d^2\implies2a\dfrac{\mathrm da}{\mathrm dt}+2b\dfrac{\mathrm db}{\mathrm dt}=0\implies\dfrac{\mathrm da}{\mathrm dt}=-\dfrac ba\dfrac{\mathrm db}{\mathrm dt}

We can solve for the angular velocity \dfrac{\mathrm d\theta}{\mathrm dt}:

\dfrac{\mathrm d\theta}{\mathrm dt}=\cos^2\theta\dfrac{b\left(-\frac ba\frac{\mathrm db}{\mathrm dt}\right)-a\frac{\mathrm db}{\mathrm dt}}{b^2}=-\dfrac1a\dfrac{\mathrm db}{\mathrm dt}

At the point when \theta=40^\circ and \dfrac{\mathrm db}{\mathrm dt}=1.8 ft/s, we get

\dfrac{\mathrm d\theta}{\mathrm dt}=-\dfrac{1.8}a\dfrac{\rm deg}{\rm s}=-\dfrac{1.8}{d\sin40^\circ}\dfrac{\rm deg}{\rm s}

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