A dinner plate, of mass 580 g is pushed 90.0cm along a dining room table by a constant force of F = 3.60 N directed at angle theta = 24.0 deg below the horizontal. If the coefficient of kinetic friction between the plate and the table's surface is 0.440 determine
a. the normal force exerted on the plate by the table
b. the work done by the applied force F
c. the work done by the force of kinetic friction
work = force * distance over which force acts * cos(theta) where theta is the angle between the displacement and force[/B]
The Attempt at a Solution
I think I understand, but my numbers seem to small to be true/actual values?
a. Sum of forces in x = Force exerted on plate in x direction (Fx) - kinetic friction (fk)
Sum of forces in y = N - w(plate) - Force exerted on plate in y direction (Fy)
Since there is no vertical acceleration, N - w - Fy = 0 --> N = w + Fy --> N = mg +Fy
where m=0.580 kg, g= 9.8, Fy= 3.60sin(24)
N= (0.58)(9.8) + 3.60sin(24) --> N = 7.15 N (this seems to small? normally normal force is in the 100s)
b) work is = Fdcos(theta) --> d=0.9 m F=3.60 N, theta is the angle between the force and displacement = 24?
Therefore, w= (3.60N)(0.9m)cos(24) --> w=2.96 J (again this seems kind of tiny for the value?)
c) Again use work
w= fk * d * cos(theta) where fk = uk*N
w=uk*N* d * cos (theta)
w= -2.83 J
I am not sure if b or c is right, but I am fairly certain for a. Could anyone verify?
Thank you in advance!