Answer to Question #144485 in Physics for bshsv

Question

Tara pushes a 10.5 kg block of ice 11.0 m across the floor with a force of 235 N. What is the ice block’s velocity the moment she stops pushing it?

Answer 1

The acceleration given to the block is (according to Newton's second law):

$$a = \dfrac{F}{m}$$

where $m = 10.5kg$ is the mass of the block and $F = 235N$ is the exerted force.

Assuming block started from rest, the distance travelled by it in time $t$ is:

$$x = \dfrac{at^2}{2}$$

As far as $x = 11m$, this time will be:

$$t = \sqrt{\dfrac{2x}{a}} = \sqrt{\dfrac{2xm}{F}}$$

The speed gained in this time is:

$$v = at = \dfrac{F}{m}\sqrt{\dfrac{2xm}{F}} = \sqrt{\dfrac{2xF}{m}}\\ v = \sqrt{\dfrac{2\cdot 11\cdot 10.5}{235}} \approx 0.99 m/s$$

Answer. 0.99 m/s.