A dinner plate, of mass 580 g is pushed .... find work/F

In summary, the conversation involved a calculation of various values related to a dinner plate being pushed along a table by a constant force. These values included the normal force exerted on the plate by the table, the work done by the applied force, and the work done by the force of kinetic friction. The calculations involved using the equations for work, force, and distance, as well as considering the angle between the displacement and force. The final values obtained were deemed to be small but were verified to be correct.
  • #1
jfnn

Homework Statement



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

Homework Equations



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=(0.440)(7.15N)(0.9m)cos(180)

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!
 
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  • #2
It all looks good to me.
 
  • Like
Likes jfnn
  • #3
TSny said:
It all looks good to me.

thank you so much.. reassurance goes a long way
 

Related to A dinner plate, of mass 580 g is pushed .... find work/F

1. What is the mass of the dinner plate?

The mass of the dinner plate is 580 grams.

2. How much force is required to push the dinner plate?

The amount of force needed to move the dinner plate depends on the friction between the plate and the surface it is being pushed on. This can be calculated using the equation F = μN, where F is the force, μ is the coefficient of friction, and N is the normal force.

3. What is the work done on the dinner plate?

The work done on the dinner plate is equal to the product of the force applied and the distance the plate is moved. This can be calculated using the equation W = Fd, where W is the work, F is the force, and d is the distance.

4. How do you find the coefficient of friction for the dinner plate?

The coefficient of friction can be found by dividing the force required to move the dinner plate by the normal force acting on the plate. This can be written as μ = F/N.

5. What is the unit for work?

The unit for work is joules (J). This is equivalent to the unit of force (newtons, N) multiplied by the unit of distance (meters, m).

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