initially four identical uniform blocks each of mass m and thickness h are spread on a table, How much work is done on the blocks in stacking them on top of one another?
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Answered by
12
I think it is mgh+2mgh+3mgh joules
noorulayn:
why not 2mgh+3mgh+4mgh?
is it wrong?
the answer written is 6mgh
but cant explain it..
yeah it willb mgh+2mgh+3mgh =6mgh because as we have to take them h height above then 2h and then 3h height.
yess i got itt thank you so much
Answered by
27
To stack them all on each other, we only need to lift 3 of them, with the remaining lowest one already being on the table.
When on the ground, a block has zero gravitational potential energy. When at a height x above the ground, the gravitational potential energy of the block is mgx. So, the work that needs to be done to lift a block from the ground to a height x is equal to mgx.
So first, we need to lift 1 block by a height h: Work done is mgh.Then, we need to lift 1 block by a height 2h: Work done is mg(2h) = 2mghFinally, we need to lift 1 block by a height 3h: Work done is mg(3h) = 3mgh
Total work done = mgh + 2mgh + 3mgh = 6mgh
i hope it will help you
When on the ground, a block has zero gravitational potential energy. When at a height x above the ground, the gravitational potential energy of the block is mgx. So, the work that needs to be done to lift a block from the ground to a height x is equal to mgx.
So first, we need to lift 1 block by a height h: Work done is mgh.Then, we need to lift 1 block by a height 2h: Work done is mg(2h) = 2mghFinally, we need to lift 1 block by a height 3h: Work done is mg(3h) = 3mgh
Total work done = mgh + 2mgh + 3mgh = 6mgh
i hope it will help you
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