Question

you push an object intially at rest across a frictionless floor with a constant force for a time interval delta t, resulting in a final speed v for the object .You repeat the experiment but with a force that is twice as large .What time interval is now required to reach the same speed (a)4delta t (b)2delta t(c)delta t/2(e)delta t/4​

Answer 1

Answer:

4 delta t

hope it helps !!!

Answer 2

Initially, intial velocity is zero, force is equal to force and final velocity is v and in second case force is 2F.

In this case we are using newtons second law

In 1st case ,

$a = \frac{v}{t}$ (u = 0)

F= ma

$a=\frac{F}{m} = \frac{v}{t}$

2nd case ,

force applied = 2F

a = will be twice as m is constant,

Time interval will be half than the first case.

Acceleration will be larger in second case.

And hence time interval will be Δ$\frac{t}{2}$