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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

Respuesta :

skyluke89
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:
[tex]\sum F = ma[/tex] 
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 [tex]F_f = \mu m g[/tex] which has opposite direction of F (because it points against the direction of the motion). Therefore we can rewrite the previous equation as
[tex]F-F_f = ma[/tex]
and solve to find F:
[tex]F=ma+F_f =m(a+\mu g)=(50 kg)(1.2 m/s^2+(0.35)(9.81 m/s^2))=[/tex]
[tex]=232 N[/tex]

Answer:

Net horizontal force, [tex]F_{net}=231.5\ N[/tex]

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, [tex]\mu=0.35[/tex]

Acceleration of the box, [tex]a=1.2\ m/s^2[/tex]

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

[tex]F_{net}=f+ma[/tex]

[tex]F_{net}=\mu mg+ma[/tex]

[tex]F_{net}=m(\mu g+a)[/tex]

[tex]F_{net}=50\times (0.35\times 9.8+1.2)[/tex]

[tex]F_{net}=231.5\ N[/tex]

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

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