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3 years ago

8

Consider lifting a box of mass m to a height h using two different methods: lifting the box directly or lifting the box using a

pulley (as in the previous part). What is Wd/Wp, the ratio of the work done lifting the box directly to the work done lifting the box with a pulley?

1 answer:

cricket20 [7]3 years ago

8 0

Answer:

Wd/wp =1

Explanation:

The work done in lifting the box directly, is the same as lifting the box using pulley.

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A tugboat tows a ship with a constant force of magnitude F1. The increase in the ship's speed during a 10 s interval is 3.0 km/h

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The magnitude of F₁ is 3.7 times of F₂

Explanation:

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Time = 10 sec

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Speed of second tugboat = 11 km/h

We need to calculate the speed

$v_{1}=\dfrac{3.0\times10^{3}}{3600}$

$v_{1}=0.833\ m/s$

The force F₁is constant acceleration is also a constant.

$F_{1}=ma_{1}$

We need to calculate the acceleration

Using formula of acceleration

$a_{1}=\dfrac{v}{t}$

$a_{1}=\dfrac{0.833}{10}$

$a_{1}=0.083\ m/s^2$

Similarly,

$F_{2}=ma_{2}$

For total force,

$F_{3}=F_{2}+F_{1}$

$ma_{3}=ma_{2}+ma_{1}$

The speed of second tugboat is

$v=\dfrac{11\times10^{3}}{3600}$

$v=3.05\ m/s$

We need to calculate total acceleration

$a_{3}=\dfrac{v}{t}$

$a_{3}=\dfrac{3.05}{10}$

$a_{3}=0.305\ m/s^2$

We need to calculate the acceleration a₂

$0.305=a_{2}+0.083$

$a_{2}=0.305-0.083$

$a_{2}=0.222\ m/s^2$

We need to calculate the factor of F₁ and F₂

Dividing force F₁ by F₂

$\dfrac{F_{1}}{F_{2}}=\dfrac{m\times0.83}{m\times0.22}$

$\dfrac{F_{1}}{F_{2}}=3.7$

$F_{1}=3.7F_{2}$

Hence, The magnitude of F₁ is 3.7 times of F₂

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