Answer:
Coefficient of friction: 0.34
Force of friction: 23.67 N
Explanation:
The free-body diagram of the body is given below.
From the above diagram, we find that
[tex]f_{\text{net}}=f_g-f_s_{}_{}_{}[/tex]
where fg is the force due to gravity and fs is the force due to friction.
Now
[tex]f_g=mg\sin 30^o[/tex]
and
[tex]f_s=\mu_sN=\mu_smg\cos 30[/tex]
Therefore, we have
[tex]f_{\text{net}}=mg\sin 30^o-\mu_smg\cos 30^o[/tex][tex]\Rightarrow f_{\text{net}}=mg(\sin 30^o-\mu_s\cos 30^o)[/tex]
Newtons second law gives fnet = ma; therefore we have
[tex]ma=mg(\sin 30^o-\mu_s\cos 30^o)[/tex]
cancelling m from both sides gives
[tex]a=g(\sin 30^o-\mu_s\cos 30^o)[/tex]
Now in our case a = 2.0 m /s^2 and g = 9.8 m /s^2 therefore,
[tex]2.0=9.8(\sin 30^o-\mu_s\cos 30^o)[/tex]
evaluating sin 30 and cos 30 gives
[tex]2.0=9.8(0.5-\frac{\mu_s\sqrt[]{3}}{2})[/tex]
solving for μ gives
[tex]\mu_s=\frac{2}{\sqrt[]{3}}(0.5-\frac{2.0}{9.8})[/tex][tex]\boxed{\mu_s=0.34}[/tex]
Now we find the force of friction from the following:
[tex]f_s=\mu_smg\sin 30[/tex]
We know that the weight of the object is 80N, meaning mg = 80, and therefore, the above equation gives
[tex]f_s=0.34\cdot80\cdot\cos 30[/tex][tex]\boxed{f_s=23.67N}[/tex]
which is our answer!
Hence, to summerise:
Coefficient of friction: 0.34
Force of friction: 23.67 N
