Coefficient of friction of box of books

[willingtolearn]

Problem 1:
A box of books weighting 300 N is shoved cross the floor of an apartment by a force of 400 N exerted downward at an angle of 35.2* below the horizontal. If the coefficient of kinetic friction between box and floor is .57, how long does it take t move the box 4 m, starting from rest ?
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w = 300 N ==> m = 30.6 kg
u (coefficient)= .57
d = 4m
v0= 0

F (f) = u time F (normal)
F (normal) is the same as F (perpendicular)
F (per) = 300 cos 35.2* = 245.1 N
F (f) = .57 time 245.1 = 139.7 N
F = ma
139.7 = 30a --> a = 4.6 m/s2
V(final) = 2ad = 36.8 m/s
From vf = vi + at --? t = 8s
What i did wrong ?



Problem 2:
A box slides down a 30* ramp with an acceleration of 1.2 m/s2. Determine the coefficient of kinetic friction between the box and the ramp.
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Can some one give me some hints ?

 

[PhanthomJay]

Problem 1:
A box of books weighting 300 N is shoved cross the floor of an apartment by a force of 400 N exerted downward at an angle of 35.2* below the horizontal. If the coefficient of kinetic friction between box and floor is .57, how long does it take t move the box 4 m, starting from rest ?
-----------------------------
w = 300 N ==> m = 30.6 kg
u (coefficient)= .57
d = 4m
v0= 0

yes.

F (f) = u time F (normal)
F (normal) is the same as F (perpendicular)

good.

F (per) = 300 cos 35.2* = 245.1 N
F (f) = .57 time 245.1 = 139.7 N

not so good. In calculating the normal force, you need to consider the weight of the books and the component of the applied 400N force in the vertical direction, and apply Newton 1.

F = ma
139.7 = 30a --> a = 4.6 m/s2

In addition to the friction force, what other force acts in the x direction? What is the net force in the x direction?

V(final) = 2ad = 36.8 m/s

this equation is wrong it should be v^2=2ad. Or just use d=1/2at^2

 

[willingtolearn]

not so good. In calculating the normal force, you need to consider the weight of the books and the component of the applied 400N force in the vertical direction, and apply Newton 1.
In addition to the friction force, what other force acts in the x direction? What is the net force in the x direction?
this equation is wrong it should be v^2=2ad. Or just use d=1/2at^2

F (parallel) = 172.9 N
Fnet = 400 - 172.9 -> 227.1 N
F = ma --> a = 7.4 m/s2
t = 1.04 s
still wrong ?

 

[PhanthomJay]

F (parallel) = 172.9 N
Fnet = 400 - 172.9 -> 227.1 N
F = ma --> a = 7.4 m/s2
t = 1.04 s
still wrong ?

In the x direction, ther are 2 forces acting: The x component of the applied 400N force, which is 400cos35.2, and the opposing friction force, which is mu times the normal force. The normal force, using Newton 1, is the book weight plus the vert component of the 400N force. .

 

[willingtolearn]

got it

 

[learningphysics]

Problem 2:
A box slides down a 30* ramp with an acceleration of 1.2 m/s2. Determine the coefficient of kinetic friction between the box and the ramp.

Divide the forces up into components perpendicular to the plane (call this y-axis) and parallel to the plane (call this x-axis)...

we already know that the normal force Fn is perpendicular to the plane...

we already know that the frictional force Ff = mu*Fn is parallel to the plane...

Now divide gravity into components parallel and perpendicular to the plane. what are the 2 components of mg...