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The work done to lift the box is equal to the force applied multiplied by the distance moved, which is 20 joules in this case (10 N * 2 m). Power is the rate at which work is done, so if the box is lifted in 1 second, the power required would be 20 watts (20 joules / 1 second).

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Q: How much power would be required to lift a 10-N box 2 meters?
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The amount of force needed to lift a petrified log is 5000 N How much work is required to lift the log 5 meters?

The work done to lift the log can be calculated using the formula: Work = Force x Distance. In this case, the force is 5000 N and the distance is 5 meters. Therefore, the work required to lift the log 5 meters would be 5000 N x 5 m = 25000 Joules.


How much work would a crane use to lift a kg car 10 meters?

The work done by a crane to lift a 1 kg car 10 meters would be 98.1 Joules. This is calculated as the product of the force required to lift the car (9.81 N/kg) and the distance the car is lifted (10 meters).


What is The amount of force needed to lift a petrified log is 5000 N How much work is re quired to lift the log 5 meters?

The work done to lift the petrified log 5 meters can be calculated using the formula: work = force × distance. In this case, the work required would be 25,000 joules (5000 N × 5 meters).


What is the minimum energy required to lift an object weighing 200 newtons to a height of 20 meters?

The minimum energy required to lift an object is equal to the work done, which is given by the formula: work = force x distance. In this case, the work done would be 200 N (force) x 20 m (distance) = 4000 joules. Therefore, the minimum energy required to lift the object weighing 200 N to a height of 20 meters is 4000 joules.


How you would halve the effort required to lift a load resting one metre from the fulcrum?

You could halve the effort required by moving the load closer to the fulcrum. Placing the load 0.5 meters from the fulcrum would reduce the effort needed to lift it. This is based on the principle of a lever, where the effort needed is inversely proportional to the distance of the load from the fulcrum.

Related questions

The amount of force needed to lift a petrified log is 5000 N How much work is required to lift the log 5 meters?

The work done to lift the log can be calculated using the formula: Work = Force x Distance. In this case, the force is 5000 N and the distance is 5 meters. Therefore, the work required to lift the log 5 meters would be 5000 N x 5 m = 25000 Joules.


How much work would a crane use to lift a kg car 10 meters?

The work done by a crane to lift a 1 kg car 10 meters would be 98.1 Joules. This is calculated as the product of the force required to lift the car (9.81 N/kg) and the distance the car is lifted (10 meters).


What is The amount of force needed to lift a petrified log is 5000 N How much work is re quired to lift the log 5 meters?

The work done to lift the petrified log 5 meters can be calculated using the formula: work = force × distance. In this case, the work required would be 25,000 joules (5000 N × 5 meters).


If it takes 3400J to lift a rock onto a ledge what power is required to lift the rock onto the ledge in 4 seconds?

The power required to lift the rock onto the ledge can be calculated as work done divided by time. In this case, it is 3400J / 4s = 850 watts. Therefore, 850 watts of power is required to lift the rock onto the ledge in 4 seconds.


What is the minimum energy required to lift an object weighing 200 newtons to a height of 20 meters?

The minimum energy required to lift an object is equal to the work done, which is given by the formula: work = force x distance. In this case, the work done would be 200 N (force) x 20 m (distance) = 4000 joules. Therefore, the minimum energy required to lift the object weighing 200 N to a height of 20 meters is 4000 joules.


How much hydrogen gas required to lift 1kg weight?

I believe it is approximately 1.13 cubic meters (1.13 litres will lift 1 gram; do the math.


How you would halve the effort required to lift a load resting one metre from the fulcrum?

You could halve the effort required by moving the load closer to the fulcrum. Placing the load 0.5 meters from the fulcrum would reduce the effort needed to lift it. This is based on the principle of a lever, where the effort needed is inversely proportional to the distance of the load from the fulcrum.


How much force to lift 50 kilograms onto a shelf 3 meters high?

To calculate the force needed to lift 50 kilograms onto a shelf 3 meters high, you would use the formula: Force = mass x gravity x height. Assuming a gravitational acceleration of 9.81 m/s^2, the force required would be approximately 1471.5 Newtons.


If a 50 crate was lifted to a height of 10 meters and it took 5 seconds to lift the crate using a machine what was the power rating of the machine?

Please use the formula for gravitational potential energy (PE = mgh) to calculate the energy required. Then divide that by the time to get the power.


If you lift 1600 meters straight up how much work did you do?

To calculate the work done when lifting an object, you can use the formula: work = force × distance. The force required to lift an object is equal to its weight, which is mass × gravity. So, you would need to know the mass of the object to calculate the work done when lifting it 1600 meters.


How much work will be needed to lift a block weighin 4 newtons and a distance of 10 meters?

The work needed to lift the block can be calculated by multiplying the force applied (4 N) by the distance it is lifted (10 m). Therefore, the work required to lift the block would be 40 joules.


What is the power rating of a machine to lift a crate in 5 seconds?

That really depends on the weight of the crate. Also, on how high you want to lift it. Calculate the energy required to lift the crate with the formula for gravitational potential energy: PE = mgh (mass x gravity x height) Then divide this by the 5 seconds to get the minimum power required. (The actual power is somewhat larger, for various reasons - the initial acceleration required, and losses due to friction.)