Answer:
a) [tex] fr =\mu N= 0.2*164.64 N= 32.928 N[/tex]
And we need a force [tex] F> 32.928 N[tex]
b) [tex] d = \frac{-v^2_i}{2a}= \frac{-(3.5m/s)^2}{-2*1.96 m/s^2}= 3.13m[/tex]
Explanation:
Part a
For this case we have the following forces illustrated on the figure attached.
If we analyze on the x axis we just have two forces, fr the friction force and F the force to mantain the movement.
So we need this condition to satisfy the movement:
[tex] F > fr[/tex]
If we analyze the forces on the y axis we have this:
[tex] \sum Fy= ma_y = 0[/tex]
Because we have constant speed for this reason the acceleration is 0
[tex] N -W = 0[/tex]
[tex] N = mg = 16.8Kg * 9.8 \frac{m}{s^2}= 164.64 N[/tex]
And by definition the friction force is defined as: [tex]fr = \mu N[/tex]
So then the friction force would be:
[tex] fr =\mu N= 0.2*164.64 N= 32.928 N[/tex]
And we need a force [tex] F> 32.928 N[/tex]
Part b
For this case we assume that F =0 and we have a friction force of 32.928 N.
From the second law of Newton we have:
[tex] F = ma[/tex]
[tex] a = \frac{F}{m}= -\frac{fr}{m}= -\frac{32.928 N}{16.8 Kg}= -1.96m/s^2[/tex]
vf =0 (final velocity, rest at the end) and vi = 3.5 m/s.
And we can find the time for the motion like this:
[tex] vf = v_i + at[/tex]
[tex] t = \frac{-v_i}{a}= \frac{-3.5 m/s}{-1.96 m/s^2}= 1.78 s[/tex]
And then we can find the distance from the following formula:
[tex] v^2_f= v^2_i + 2a d[/tex]
[tex] d = \frac{-v^2_i}{2a}= \frac{-(3.5m/s)^2}{-2*1.96 m/s^2}= 3.13m[/tex]
