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conorwood
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Homework Statement
A box of mass = 20 kg rests on top of a cart mass = 30 kg. A force of 200 N is applied to the cart, and the box starts to slide. If the coefficient of kinetic friction is 0.25, what is the time it takes for the box to slide 1.5 m. Both the cart and the box start from rest.
m1 = 20 kg
m2 = 30 kg
F = 200 N
μk = 0.25
Δs = 1.5 m
t = ?
Homework Equations
Ffriction = (μk)(N)
Fnet = ma
aa/b = ab - aa
Δs (from rest) = (1/2)(a)(t2)
The Attempt at a Solution
Here is my free body diagram of the box:
-----------------------------[itex]\uparrow[/itex] N1
--------------------|--------------------|
--------------------|--------------------|
--------------------|--------------------|
--------------------|--------------------|
--------------------|--------------------|
---------------------------[itex]\downarrow[/itex]m1g ----- [itex]\rightarrow[/itex]Ffriction (FF)
y
|__x
Here is the kinetic diagram:
--------------------|--------------------|
--------------------|--------------------|
--------------------|--------------------| [itex]\rightarrow[/itex] m1a1
--------------------|--------------------|
--------------------|--------------------|
And the force analysis:
[itex]\Sigma[/itex] Fy = m1a1y = 0; 0 = N1 - m1g ; N1 = m1g = (20)(9.81) = 196.20 N
m1g = 196.20 N [itex]\downarrow[/itex]
N1 = 196.20 N [itex]\uparrow[/itex]
FF = (N1)(μk) = (196.2)(0.25) = 49.05 N [itex]\rightarrow[/itex]
[itex]\Sigma[/itex] Fx = m1a1 = 20a1; FF= 20a1; 49.05 = 20a1
a1 = (49.05)/(20) = 2.4525 m/s2 [itex]\rightarrow[/itex]
Here is my FBD of the cart:
-------------------- [itex]\uparrow[/itex] m1g --- [itex]\uparrow[/itex] N2 --- [itex]\leftarrow[/itex]FF
--------------------|--------------------|
--------------------|--------------------|
--------------------|--------------------| [itex]\rightarrow[/itex] F (200 N)
--------------------|--------------------|
--------------------|--------------------|
---------------------------- [itex]\downarrow[/itex]m2g
y
|__x
Here is the kinetic diagram:
--------------------|--------------------|
--------------------|--------------------|
--------------------|--------------------| [itex]\rightarrow[/itex] ma2
--------------------|--------------------|
--------------------|--------------------|
Here is the force analysis:
ƩFx = ma2; 200 - FF = ma2
200 - 49.05 = 150.95 = ma2
And here, is the problem I am having troubles understanding:
I am not sure what mass to insert here; 30 kg (the mass of the cart) or 50 kg (the mass of the cart and the box)
The solutions book puts in 30 kg, and doesn't say why it neglects the mass of the box, which would make the total mass 50 kg. Is the solutions book correct or does it make an error here? If it is correct, can somebody explain what exactly is happening and why you can do this?
putting in the 30 kg (the mass of the cart) gives an acceleration of 5.03166.
subtracting 2.4525 (the acceleration of the box) gives a relative acceleration of 2.579166
Using the kinetic formula the time for the box to move 1.5 meters gives a solution for time of about 1.08 seconds.
on the other hand, putting in 50 kg (the mass of the cart and the box) gives an acceleration of 4 and subtracting 2.4525 gives a relative acceleration of 1.5475. the kinetic formula then spits out an answer of t = 1.39 seconds.
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