A 50-kg box is being pushed along a horizontal surface. The coefficient of static friction between the box and the ground is 0.65, and the coefficient of kinetic friction is 0.35. What horizontal force must be exerted on the box for it to accelerate at 1.2 m/s2?
A) 60 N
B) 230 N
C) 120 N
D) 170 N
E) 490 N

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cjmejiab cjmejiab · Answer 1 · 2019-11-23T04:54:42+01:00

To solve the exercise it is necessary to apply the concepts related to Newton's Second Law, as well as the definition of Weight and Friction Force.

According to the problem there is a movement in the body and it is necessary to make a sum of forces on it, so that

[tex]\sum F = ma[/tex]

There are two forces acting on the body, the Force that is pushing and the opposing force that is that of friction, that is

[tex]F - F_f = ma[/tex]

To find the required force then,

[tex]F=F_f+ma[/tex]

By definition we know that the friction force is equal to the multiplication between the friction coefficient and the weight, that is to say

[tex]F = \mu mg +ma[/tex]

[tex]F = 0.35*50*0.8+50*1.2[/tex]

[tex]F=(171.5N)+(50Kg)(1.2m/s^2)[/tex]

[tex]F=231.5N[/tex]

[tex]F\approx 230N[/tex]

Therefore the horizontal force applied on the block is B) 230N