Question

a 50kg box is being pushed along a horizontal surface. the coefficient of kinetic friction between the box and the ground is 0.35.what horizontal force must be exerted on the box for it to accelerate at 1.20m/s^2

Answer 1

For Newton's second law, the resultant of the forces acting on the box is equal to the product between the mass of the box m and its acceleration a:
$\sum F = ma$ 
We are interested only in what happens on the x-axis (horizontal direction). Only two forces act on the box in this direction: the force F, pushing the box along the surface, and the frictional force $F_f = \mu m g$ which has opposite direction of F (because it points against the direction of the motion). Therefore we can rewrite the previous equation as
$F-F_f = ma$
and solve to find F:
$F=ma+F_f =m(a+\mu g)=(50 kg)(1.2 m/s^2+(0.35)(9.81 m/s^2))=$
$=232 N$

Answer 2

Answer:

Net horizontal force, $F_{net}=231.5\ N$

Explanation:

It is given that,

Mass of the box, m = 50 kg

The coefficient of kinetic friction between the box and the ground is 0.35, $\mu=0.35$

Acceleration of the box, $a=1.2\ m/s^2$

We know that the frictional force acts in opposite direction to the direction of motion. The net force acting on it is given by :

$F_{net}=f+ma$

$F_{net}=\mu mg+ma$

$F_{net}=m(\mu g+a)$

$F_{net}=50\times (0.35\times 9.8+1.2)$

$F_{net}=231.5\ N$

So, the net force acting on the box is 231.5 N. hence, this is the required solution.