A box slides on top of a cart as a force is applied to the cart

[conorwood]

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
μ_k = 0.25
Δs = 1.5 m
t = ?

Homework Equations

F_friction = (μ_k)(N)

F_net = ma

a_a/b = a_b - a_a

Δs (from rest) = (1/2)(a)(t²)

The Attempt at a Solution

Here is my free body diagram of the box:

-----------------------------\uparrow N₁
--------------------|--------------------|
--------------------|--------------------|
--------------------|--------------------|
--------------------|--------------------|
--------------------|--------------------|
---------------------------\downarrowm₁g ----- \rightarrowF_friction (F_F)

y
|__x

Here is the kinetic diagram:

--------------------|--------------------|
--------------------|--------------------|
--------------------|--------------------| \rightarrow m₁a₁
--------------------|--------------------|
--------------------|--------------------|


And the force analysis:

\Sigma Fy = m₁a_1y = 0; 0 = N₁ - m₁g ; N₁ = m₁g = (20)(9.81) = 196.20 N

m₁g = 196.20 N \downarrow
N₁ = 196.20 N \uparrow

F_F = (N₁)(μ_k) = (196.2)(0.25) = 49.05 N \rightarrow

\Sigma Fx = m₁a₁ = 20a₁; F_F= 20a₁; 49.05 = 20a₁

a₁ = (49.05)/(20) = 2.4525 m/s² \rightarrow


Here is my FBD of the cart:

-------------------- \uparrow m₁g --- \uparrow N₂ --- \leftarrowF_F
--------------------|--------------------|
--------------------|--------------------|
--------------------|--------------------| \rightarrow F (200 N)
--------------------|--------------------|
--------------------|--------------------|
---------------------------- \downarrowm₂g

y
|__x

Here is the kinetic diagram:

--------------------|--------------------|
--------------------|--------------------|
--------------------|--------------------| \rightarrow ma₂
--------------------|--------------------|
--------------------|--------------------|


Here is the force analysis:

ƩFx = ma₂; 200 - F_F = 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.

 

[ehild]

ƩFx = ma₂; 200 - F_F = 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)

When you drew the free body diagram you considered both object separately. The resultant of all forces acting on one object is the acceleration of that object multiplied by the mass of the same object.



ehild