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Bingel [31]
2 years ago
7

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 o

f the ramp slows the object until it comes to a stop, how much work must the factory worker have done
Physics
1 answer:
ruslelena [56]2 years ago
8 0

Answer:

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

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

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egoroff_w [7]

Answer:

The resultant force would (still) be zero.

Explanation:

Before the 600-N force is removed, the crate is not moving (relative to the surface.) Its velocity would be zero. Since its velocity isn't changing, its acceleration would also be zero.

In effect, the 600-N force to the left and 200-N force to the right combines and acts like a 400-N force to the left.

By Newton's Second Law, the resultant force on the crate would be zero. As a result, friction (the only other horizontal force on the crate) should balance that 400-N force. In this case, the friction should act in the opposite direction with a size of 400 N.

When the 600-N force is removed, there would only be two horizontal forces on the crate: the 200-N force to the right, and friction. The maximum friction possible must be at least 200 N such that the resultant force would still be zero. In this case, the static friction coefficient isn't known. As a result, it won't be possible to find the exact value of the maximum friction on the crate.

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3 years ago
A wooden log is displaced to a distance of 20m in 10 seconds by applying 500N effort . Calculate the workdone and power.​
nata0808 [166]

A wooden log is displaced to a distance of 20m in 10 seconds by applying 500N effort . Calculate the workdone and power...

Solution,

displacement = 20 m

time = 10 sec

force = 500 N

work done = ?

power = ?

Now ,

work done = f × s

= 500 N × 20 m

= 10000 j

Now ,

power \:  =  \frac{w}{t}  \\  =  \frac{10000j}{10sec}  \\   = 1000W

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Answer:

Explanation:

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