Answer:
a)[tex]f_k=172.17N[/tex]
b)[tex]a_x=0.428m/s^2[/tex]
Explanation:
The magnitude of the kinetic frictional force is given by the formula
[tex]f_k=\mu_k N[/tex]
In this case, on the vertical direction the only forces are the weight and the normal force, and since the object does not move in that direction they must be equal. We have then:
[tex]f_k=\mu_k N=\mu_k W = \mu_k mg[/tex]
Substituting values we get
[tex]f_k=\mu_k mg=(0.27)(65Kg)(9.81m/s^2)=172.17N[/tex]
On the horizontal direction the only forces are the one the person exerts ([tex]F_p[/tex]) and the kinetic friction opposing it, so the net force is:
[tex]F_x=F_p-f_k[/tex]
We apply then Newton's 2nd Law to calculate the acceleration of the object in the horizontal direction:
[tex]ma_x=F_x=F_p-f_k[/tex]
[tex]a_x=\frac{F_p-f_k}{m}[/tex]
And substituting values we have
[tex]a_x=\frac{200N-172.17N}{65Kg}=0.428m/s^2[/tex]
