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

Net horizontal force,
`F_(net)=231.5\ N`

Step-by-step 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* (0.35* 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.