Question

A 1.2 kg box is being pushed horizontally on a smooth
surface by a 10 N force for 6 m. Given that the box is
initially at rest, what is the speed of the box at the end of
the 6 m?​

Answer 1

Answer:

Explanation:

Given that force = 10N

F= ma

10 = 1.2 a

a = 100/12

V^2 = u^2 + 2as

V^2 = 2×100× 6/12

V = 10 m/s

Answer 2

Explanation:

F=MA (mass* acceleration)

MASS OF THE BOX= 1.2 KG

FORCE ACTING ON THE BOX= 10 N(NEWTONS)

ACCELERATION= FINAL VELOCITY - INITIAL VELOCITY/TIME

FINAL VELOCITY= 6METRES

INITIAL VELOCITY=0 METRES

TIME=6MINUTES

SO a= 6m-0m/6min=1

so acceleration = 1m/sec square

speed= distance/time

6metres/6meters

=1 metre/sec