Question

Elle is pushing a box of mass 7.0 kilograms with the force of 25 newtons. If the force of friction is 2.6 newtons, what is the value of acceleration of the box? 0.7 meters/second 2.6 meters/second 3.2 meters/second 3.6 meters/second

Answer 1

F=mxa
(25-2.6)=7 X a
22.4\7=a
a =3.2 metres per second squared

Answer 2

Answer:

Acceleration, $a=3.2\ m/s^2$

Explanation:

It is given that,

Mass of the box, m = 7 kg

Force acting on the box, F = 25 N

Frictional force, f = 2.6 N

We need to find the acceleration of the box. The net force acting on the box is ( F - f ) = (25 - 2.6) = 22.4 N

Using second law of motion as :

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

$a=\dfrac{22.4\ N}{7\ kg}$

$a=3.2\ m/s^2$

So, the acceleration of the box is $3.2\ m/s^2$. Hence, this is the required solution.