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.