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rusak2 [61]
4 years ago
5

A body builder loads a bar with 550 newtons (~125 pounds) of weight and pushes the bar over her head 10 times. Each time she lif

ts the weight 0.5 meters. How much work did she do? 275 Joules. If she lifts the bar 10 times in 20 seconds, how much power did she use?
Physics
1 answer:
Neporo4naja [7]4 years ago
6 0

Work = (force) x (distance)

 (550 newtons) x (0.5 meter) = 275 joules each lift .

 (275 joules/lift) x (10 lifts)  =  2,750 joules of work all together.


Power = (work done) / (time to do the work)

           = (2,750 joules) / (20 seconds)

           =    137.5 watts .        (about 0.18 horsepower)

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Answer:

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(b) 11000 N

Explanation:

Note that answers are given to 2 significant figures which is what we have in the values in the question.

(a) Speed is given by the ratio of distance to time. In the question, the time given was the time it took the pulse to travel the length of the cable twice. Thus, the distance travelled is twice the length of the cable.

v=\dfrac{2\times 620 \text{ m}}{14\text{ s}} = \dfrac{1240\text{ m}}{14\text{ s}}=88.571428\ldots \text{ m/s}= 89\text{ m/s}

(b) The tension, T, is given by

v =\sqrt{\dfrac{T}{\mu}}

where v is the speed, T is the tension and \mu is the mass per unit length.

Hence,

T = \mu\cdot v^{2}

To determine \mu, we need to know the mass of the cable. We use the density formula:

\rho = \dfrac{m}{V}

where m is the mass and V is the volume.

m=\rho\cdot V

If the length is denoted by l, then

\mu = \dfrac{m}{l} = \dfrac{\rho\cdot V}{l}

T = \dfrac{\rho\cdot V}{l} v^{2}

The density of steel = 8050 kg/m3

The cable is approximately a cylinder with diameter 1.5 cm and length or height of 620 m. Its volume is

V = \pi \dfrac{d^{2}}{4} l

T = \dfrac{\rho\cdot\pi d^2 l}{4l}v^2 = \dfrac{\rho\cdot\pi d^2}{4}v^2

T = \dfrac{8050\times\pi\times0.015^2}{4} \times 88.57^2

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Answer:

a)

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The limit for falling off would be N_{top}=0, so the minimum speed would be:

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