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An object with a mass of 10 kg is rolled down a frictionless ramp from a height of 3 meters. If a factory worker at the bottom of the ramp slows the object until it comes to a stop, how much work must the factory worker have done

User HGPB
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Answer:

The amount of work the factory worker must to stop the rolling ramp is 294 joules

Step-by-step explanation:

The object rolling down the frictionless ramp has the following parameters;

The mass of the object = 10 kg

The height from which the object is rolled = 3 meters

The work done by the factory worker to stop the rolling ramp = The initial potential energy, P.E., of the ramp

Where;

The potential energy P.E. = m × g × h

m = The mass of the ramp = 10 kg

g = The acceleration due to gravity = 9.8 m/s²

h = The height from which the object rolls down = 3 m

Therefore, we have;

P.E. = 10 kg × 9.8 m/s² × 3 m = 294 Joules

The work done by the factory worker to stop the rolling ramp = P.E. = 294 joules

User GianMS
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