The 2.0 kg wood box in the figure(Figure 1) slides down a vertical wood wall while you push on it at a 45 ∘ angle. The coefficient of kinetic friction of wood on wood is μk = 0.200.

Question

What magnitude of force should you apply to cause the box to slide down at a constant speed?

Attempt:

Fnet(y) = Fp(y) - (2.0)(9.8) = 0
Fp = N (μk)
Fp = N (0.200)
Fp = (9.8)(2.0)(0.200)
Fp = 3.92

Fp(y) = 3.92 (Sin(45))
Fp(y) = 2.77

Fnet(y) = 2.77 - (2.0)(9.8)
Fnet(y) = 16 N

The answer is supposed to be 23. What am I doing wrong?

Answer 1

If I read the problem, you are also pushing upwards Fcos45, as well as inwards.
in the vertical:
mg-Fcos45-mu*Fsin45=O
2*9.8-F(.707-.2*.707)=0
F= 19.6/(0.5656)=34.7
Now if you are pushing downward at 45deg, then
2*9.8 F(-.707-.2*.707)=0
F=19.6/( 0.848)=23N
so it I guess has the drawing with the force downward

Answer 2

a velocity of magnitude 40m/s is directed at angel of 40 degree east of north draw a vector

Answer 3

This question is so confusing still being assigned in 2022

Answer 4

Constant Velocity = 0m/s^2
So:

Fnety = Fpy+Fk-Fg = 0

Fpy is the push upwards and because you want the box to travel downwards Fk is positive.

Fpsin(45) + 0.2Fn - Fg = 0

We do not have Fn

Fnetx = Fn-Fpx = 0
Fn = Fpcos(45)

Substitute:

Fnety = Fpcos(45) + 0.2Fpsin(45) - Fg = 0

Simplify:
Fp(cos(45) - 0.2sin(45)) = Fg

Fp = Fg/(cos(45) + 0.2sin(45))

Fp = (9.8*2)/(cos(45) + 0.2sin(45))

Fp = 23N