A 6.0kg block is released from rest 80m above the ground. When it has fallen 60m its kinetic energy is:

Question

I got 1200J as my answer
mgh=(6.0*9.81* 60m)=3531.6
PE + KE= (ME + KE)
3531.6 + KE= (6.0*9.81*80m)
KE= 1177.2
is that right

A 6.0 Kg block is thrown up from a point 20m above Earth's surface. At what heights above Earth's surface will the gravitational potential energy of the earth-block system have increase by 500J.
mgh= 2(9.81)(20m)= 392.4 --> potential
500J= mgh
500J = (2kg)(9.81) (h)
h= 25m
someone had told me my actual answer for height should be 45m, because I need to add the 20 I started with.

Thank you

The KE falling sixty meters is 3500J, as you calculated It lost that much gravitational PE, so it is transformed to KE.
If you wnat to do it this way
PE + KE= (ME + KE)
PE(at20) + KE(at20)= (6.0*9.81*80m)
mg20 +KE(at20)= (6.0*9.81*80m)
and solve for KE at 20 , it will be the same as above.

Second question.
mg(distanceup )=500
YOu do not add the 20, because the problem is dealing with INCREASE in PE, not total PE.

Answer 1

So
2(9.81)(h)=500
h= 25m