Simple Work Energy Principle Question

[Kathy W]

I think I did most of this question correctly, but I think I might be overlooking a few steps because I might have thought about it too simply.

1. A dinner plate of mass 510 g is pushed 60 cm along a dining table by a constant force of 3.0 N directed 22° below the horizontal. If the coefficient of kinetic friction between the plate and the table’s surface is 0.44, determine the work done on the plate by
   • a) the applied force
   • b) the force of gravity
   • c) the normal force
   • d) the force of kinetic friction

Attempt at solution

a) Work of FApplied= Fdcosθ
= (3N)(0.6m)(cos22)
=1.67 J

Am I correct in assuming this was asking about the horizontal work because there was no movement in the y-plane?

b) Force of Gravity
Fgravity=mg
=(0.51)(9.8)
=4.998 N

Work gravity= Fd
= (4.998N)(0)
=0 Joules

c) **This is where I am also confused... the normal force is greater than the force of gravity do to the additional force being applied downward, does that mean that work is done in this plane or is it still 0 because there is no movement?

Fapplied y= (sin22)(3)
FAy= 1.124

FN=mg+FAy
= 4.998+1.124
= 6.122 N

W= Fd
= (6.122)(0)
= 0 Joules ***? Is this right, I'm really not sure?

d) Force of kinetic friction

Ffr= μkFN
=(0.44)(6.122)
=2.69N

Work of friction= Fdcosθ
= (2.69)(0.6)(cos180)
= -1.62 Joules

I know this is a simple question, I just want to make sure I am in the correct direction with my post, I appreciate any help immensely!

 

[Chandra Prayaga]

Since this looks like a homework question, I will ask: What makes you doubt your answer for part c)?

 

[Qwertywerty]

Yes , the work done by normal would be zero ( N ⊥ d ) .

I think the rest is fine .