Question

A 0.60-kg block initially at rest on a frictionless horizontal surface is acted upon by a force of 4.0 N for a distance of 6.5 m. How much kinetic energy does the block gain?

A 0.60-kg block initially at rest on a frictionless horizontal surface is acted upon by a force of 4.0 N for a distance of 6.5 m. How much kinetic energy does the block gain?

Answer 1

`"26 Joule"`

Explanation:

According to the work-energy theorem, the net work on an object causes a change in the kinetic energy of the object.

Change in Kinetic energy = Work done on object

`∆"K.E" = F × d = "4.0 N" × "6.5 m" = "26 Joule"`

Therefore, block gains `"26 Joule"` of Kinetic energy.

Answer 2

acceleration of the block was `F/m=4/0.60=6.67 ms^-2`

So,initially it was at rest,and its kinetic energy was `1/2m 0^2=0`

Now,after going `6.5m` if it achieved a velocity of `v` then,

`v^2=0^2 +2*6.67*6.5=86.71` (using, `v^2=u^2+2as`)

So,after travelling that much distance its kinetic energy became, `1/2 0.60 *86.71=26.013 J`

So,gain in kinetic energy = `26.013-0=26.013 J`