Energy Calculations please check work (thanks)

[blackout85]

Please someone check to see if on the right track:

A woman lifts a barbell 2.0m in 5.0s, beginning and ending at rest. If she lifts the same distance in 10s, the work done by her is:

I said it would be four times as much because I used 1/2mv^2. I placed the v I calculated from v=d/t into the kinetic equation. (.4)^2 (.2)^2
.16 is four times as much as .04. Is this right.

A 2kg block is thrown upward from a point 20m above the Earth's surface. At what height above the Earth's surface will the gravitational potential energy of the earth-block systme have increased by 500J.

work:

mgh= 2(9.81)(20m)= 392.4 --> potential
500J= mgh
500J = (2kg)(9.81) (h)
h= 25m

If someone could point me in the right direction of how to do this problem I would appreciate it. Thank you.

 

[rsk]

Please someone check to see if on the right track:

A woman lifts a barbell 2.0m in 5.0s, beginning and ending at rest. If she lifts the same distance in 10s, the work done by her is:

I said it would be four times as much because I used 1/2mv^2. I placed the v I calculated from v=d/t into the kinetic equation. (.4)^2 (.2)^2
.16 is four times as much as .04. Is this right.

I don't think so - I think the work will be the same - Work Done = Force x distance - taking longer means less power but the same work.

A 2kg block is thrown upward from a point 20m above the Earth's surface. At what height above the Earth's surface will the gravitational potential energy of the earth-block systme have increased by 500J.

work:

mgh= 2(9.81)(20m)= 392.4 --> potential
500J= mgh
500J = (2kg)(9.81) (h)
h= 25m

If someone could point me in the right direction of how to do this problem I would appreciate it. Thank you.

This one I think is correct - except the actual height is 45, you need to add the 20m you started with.. When you're as close to the Earth as this, the mgh formula is fine.

 

[kyle8921]

To check if you are on the right track, let's break down the problem into smaller steps and use the correct equations for each step.

First, let's calculate the work done by the woman in the first scenario. We can use the equation W = Fd, where W is work, F is force, and d is distance. We know that the woman lifted the barbell 2.0m, so we need to find the force she used. We can use the equation F = ma, where m is the mass of the barbell and a is the acceleration. Since the barbell was initially at rest and ended at rest, the acceleration is equal to 0. Therefore, the force used is also equal to 0. This means that the work done in the first scenario is also equal to 0.

Next, let's calculate the work done in the second scenario. We already know that the distance is 2.0m, so we just need to find the force used. Using the same equation as before, F = ma, we can find that the force used is equal to 1/2mv^2. Plugging in the values of 2kg for mass and 0.4m/s for velocity (calculated by dividing 2m by 5s), we get a force of 0.4N. Now, we can calculate the work done using the equation W = Fd. Plugging in the values of 0.4N for force and 2.0m for distance, we get a work of 0.8J.

To check if this answer is four times as much as the work done in the first scenario, we can compare it to the work done in the first scenario (which we found to be 0J). Since 0.8J is four times greater than 0J, this means that your answer is correct!

For the second problem, we are given the gravitational potential energy of the earth-block system (500J) and asked to find the height at which this energy is reached. To do this, we can use the equation for gravitational potential energy, U = mgh, where U is the potential energy, m is the mass, g is the acceleration due to gravity, and h is the height.

We already know the values for m (2kg) and g (9.81m/s^2), so we can rearrange the equation to solve