Question

A block resting on a smooth horizontal floor is being pushes by a constant horizontal force. If the kinetic energy gained by the block in the 1st, 2nd and the 3rd second is K1, K2 and K3. Find K1:K2:K3

Answer 1

from conservation of energy theorem loss is equel to gain in absence of non conservative forces

Answer 2

Given:

• a block resting on a smooth horizontal floor is being pushed by a constant horizontal force.
• the kinetic energy gained by the block in the 1st, 2nd, and 3rd second is k1, k2, and k3.

To find: Ratio between R1:R2:R3?

let's constant force = f

mass of block is      = m

                           K1 : K2 : K3

W. D. by f in 1st sec. 2nd sec. and 3rd sec. will be,

                           fd1 : fd2 : fd3

Now,

                  $d_{1} = \frac{1}{2} at^{2}$

                       = $\frac{1}{2} (\frac{f}{m} .(1^{2})$

The same will go with d2 and d3, only the value of time will change at the place of " t "

                  $d2= \frac{1}{2} \frac{f}{m}.(2^{2}) \\d3= \frac{1}{2} \frac{f}{m}.(3^{2})$

            $f[\frac{1}{2} (\frac{f}{m} .(1^{2})] : f[\frac{1}{2} (\frac{f}{m} .(2^{2})] : \frac{1}{2} (\frac{f}{m} .(3^{2})]$

                        $(1^{2})$  : $(2^{2})$  : $(3^{2} )$

                            1 : 4 : 9

Hence the ratio between k1:k2:k3 is 1:4:9.