two blocks A and B of same mads are released from top of two different smooth inclined planes. if the height of both the inclined plane is same then the work done by gravity by the time both blocks reach the bottom of respective inclines is
Answers
Answer:
∴ work done by gravity by the time both blocks reach the bottom of respective inclines is SAME = mgΔh
Explanation: Let the height be 'H' and mass of each block be 'M'. These quantities remain same in both the cases.
Let the inclination be θ.
the block is being displaced in two directions(x and y).
Let, it was at initially at (α, β). Let our reference be ground(surface).
So, β = position in y-axis = height = H
Hence, it is at (α. h). [initially]
When it reaches the bottom, let its position be (γ, 0) [finally]
Displacement = (γ - α, h - 0) = (γ - α, 0 - h)
which is represented by (γ - α)i - hj.
The acting force is gravity(downwards, in -j) = mg(-j) = - mg j
Therefore, by the basic definition of work-done,
work by gravity = force * displacement * cosΘ
= - mg(j) * [(γ - α)i - hj]
= - mg * (γ - α) ij + mghi²
= 0 + mgh [ij = 0, i² = 1]
= mgh
Which means, workdone by gravity is independent of the horizontal position(displacement) and angle of inclination. It just depends on the vertical positions.
∴ If bodies have same initial and final vertical positions, work-done by gravity on the bodies is same. [irrespective of anything else (in all the normal cases) ]
∴ work done by gravity by the time both blocks reach the bottom of respective inclines is SAME = mgΔh