Physics help! Please!

Question

A constant horizontal force of 5 N is required to push a box along a horizontal floor at a constant velocity of 4 m/s. What constant horizontal force is needed if you wish to push this box at a constant acceleration of 4 m/s^2? Assume the coefficient of kinetic friction between the box and the floor is 0.7.

Thanks to anyone who helps!

Answer 1

First of all we need to calculate the mass of the box .

We know that coefficient of friction is 0.7

Friction = µR

When box was moving at constant velocity , friction was equal to 5 N .

5 =0.7R

R = 50/7 = mg

9.81m = 50/7

m ≈ 0.728 Kg

Force - 0.7(50/7) = 0.728(4)

Force = 7.912 N ≈ 8 N

Answer 2

The force of friction must be f=5 N

f= u mg  where u = 0.7

m= f/(ug)  

Total horizontal force is then

Ft= ma + f

Ft=[f/(ug) ]a + f

Ft= (5/(0.7 x 9.81)] 4 + 5

Ft= 2.9 + 5= 7.9N

Answer 3

From the first statement, since a=0, we have

F=5N = cfr*Weight of block=.7*W

That means W = 5/.7=7.14N

Now we can use F-cfr*W = ma

F = (W/g)*a + cfr*W = W(cfr+a/g) = 7.14(.7+4/9.8) =7.91 N

g = gravity =9.8 m/s^2

The answer is 7.91 N