Question

a block of mass M is released on a smooth inclined plane from a height h the kinetic energy of block on reaching the ground is​

Answer 1

Answer:

v² = u² + 2gh

v² = 0 + 2gh

so v on reaching ground is ,

v = √2gh

So , KE = 1/2 mv²

= 1/2 M (√2gh)²

= 1/2 M × 2gh

= Mgh

Answer 2

The kinetic energy of block on reaching the ground is Mgh.

Explanation:

Given that,

Mass of block = M

Height = h

We need to calculate the kinetic energy of block

Using conservation of energy

$K.E_{t}+P.E_{t}=K.E_{b}+P.E_{b}$

The kinetic energy of the block at top is zero.

The potential energy of the block at bottom is zero.

$0+P.E_{t}=K.E_{b}+0$

$K.E_{b}=mgh$

Put the value into the formula

$K.E_{b}=Mgh$

Hence, The kinetic energy of block on reaching the ground is Mgh.

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Topic : Kinetic energy

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