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2 years ago

an object has a mass of 50kg, a final height of 20m and an initial height of 8m. what is the amount of work done

Physics

1 answer:

Andrei [34K]2 years ago

7 0

amount of work done is 5880 J

Given:

mass of object = 50kg

Final height = 20m

initial height = 8m

To Find:

amount of work done

Solution:

work is done when a force acts upon an object to cause a displacement. You can calculate the energy transferred, or work done, by multiplying the force by the distance moved in the direction of the force.

The work done by gravity is given by the formula,

W = mgh

W = 50 x 9.8 x ( 20-8)

= 5880 J

So the work done is 5880 J

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Two workers are sliding 300 kg crate across the floor. One worker pushes forward on the crate with a force of 400 N while the ot

almond37 [142]

Answer:

The kinetic coefficient of friction of the crate is 0.235.

Explanation:

As a first step, we need to construct a free body diagram for the crate, which is included below as attachment. Let supposed that forces exerted on the crate by both workers are in the positive direction. According to the Newton's First Law, a body is unable to change its state of motion when it is at rest or moves uniformly (at constant velocity). In consequence, magnitud of friction force must be equal to the sum of the two external forces. The equations of equilibrium of the crate are:

$\Sigma F_{x} = P+T-\mu_{k}\cdot N = 0$ (Ec. 1)

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Where:

$P$ - Pushing force, measured in newtons.

$T$ - Tension, measured in newtons.

$\mu_{k}$ - Coefficient of kinetic friction, dimensionless.

$N$ - Normal force, measured in newtons.

$W$ - Weight of the crate, measured in newtons.

The system of equations is now reduced by algebraic means:

$P+T -\mu_{k}\cdot W = 0$

And we finally clear the coefficient of kinetic friction and apply the definition of weight:

$\mu_{k} =\frac{P+T}{m\cdot g}$

If we know that $P = 400\,N$ , $T = 290\,N$ , $m = 300\,kg$ and $g = 9.807\,\frac{m}{s^{2}}$ , then:

$\mu_{k} = \frac{400\,N+290\,N}{(300\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}$

$\mu_{k} = 0.235$

The kinetic coefficient of friction of the crate is 0.235.

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Explanation:

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