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Mila [183]
2 years ago
13

Calculate the work done if a boy lifts a bag of cement 500N to the floor of a lorry 2.5m above the ground

Physics
1 answer:
eduard2 years ago
4 0

Answer:

W = 1250 J = 1.25 KJ

Explanation:

The work done by the boy is due to the change in the position of the cement vertically. Hence, the work done in this case will be equal to the potential energy of the cement:

W = P.E\\W = mgh

where,

W = Work done = ?

mg = W = weight of cement = 500 N

h = height covered = 2.5 m

Therefore,

W = (500\ N)(2.5\ m)

<u>W = 1250 J = 1.25 KJ</u>

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ILI
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The original kinetic energy will be 0 J and the final kinetic energy will be 7500 J and the amount of work utilized will be similar to the final kinetic energy i.e., 7500 J.

<u>Explanation:</u>

As it is known that the kinetic energy is defined as the energy exhibited by the moving objects. So the kinetic energy is equal to the product of mass and square of the velocity attained by the car. Thus,

                  \text {Kinetic energy}=\frac{1}{2} m v^{2}

So the initial kinetic energy will be the energy exerted by the car at the initial state when the initial velocity is zero. Thus the initial kinetic energy will be zero.  

The final kinetic energy is

\text {Kinetic energy}=\frac{1}{2} m v^{2}=\frac{1}{2} \times 600 \times 5 \times 5 = 7500 J

As the work done is the energy required to start the car from zero velocity to 5 m/s velocity.  

                       Work done = Final Kinetic energy - Initial Kinetic energy

Thus the work utilized for moving the car is  

                         Work done = 7500 J - 0 J = 7500 J

Thus, the initial kinetic energy of the car is zero, the final kinetic energy is 7500 J and the work utilized by the car is also 7500 J.

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