Friction problem (pushing a box)

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The discussion revolves around calculating the acceleration of a 60 kg box being pushed with a net external force of 90 N while facing a frictional force of 30 N. Two different solutions were proposed, leading to confusion about the correct approach. The first solution correctly uses the net force to find the acceleration as 1.5 m/s², while the second solution mistakenly calculates the total force without accounting for friction, yielding an incorrect acceleration of 2.0 m/s². Participants agree that the problem's wording is misleading, as it introduces friction but provides the net force directly. Ultimately, the consensus is that the first solution is the correct interpretation of the problem.
Alyona
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Homework Statement


A person is pushing a box. The net external force on the 60 kg box is stated to be 90 N. If the force of friction opposing the motion is 30 N, what is the acceleration of the box?

Homework Equations


This seems to be a pretty easy, straightforward problem, but I'm confused. Is the first or second solution correct?

The Attempt at a Solution


Fnet = F - f = ma
90 N = F - 30 N = (60 kg)a
a = 90 N / 60 kg = 1.5 m/s2

Fnet= F - f = ma
90 N = F - 30 N
F = 120 N
F = ma
120 N = (60 kg)a
a = 120 N / 60 kg = 2.0 m/s2
 
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Alyona said:

Homework Statement


A person is pushing a box. The net external force on the 60 kg box is stated to be 90 N. If the force of friction opposing the motion is 30 N, what is the acceleration of the box?

Homework Equations


This seems to be a pretty easy, straightforward problem, but I'm confused. Is the first or second solution correct?

The Attempt at a Solution


Fnet = F - f = ma
90 N = F - 30 N = (60 kg)a
a = 90 N / 60 kg = 1.5 m/s2

Fnet= F - f = ma
90 N = F - 30 N
F = 120 N
F = ma
120 N = (60 kg)a
a = 120 N / 60 kg = 2.0 m/s2
I am confused, too. The problem statement says the net external force is 90N and then introduces another external force - friction. Friction certainly an external force - it is a force the surface exerts on the box. So I think your first answer is correct given the way the problem is stated, but I am not sure that was the intent of the person who wrote the problem.
Edit: I do not think your second answer is correct - that answer ignores the force of friction on box.
 
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Yes, it's confusing. If you don't need the force of friction to solve the problem, since the net force was given, why put it in the problem then? I think the first solution is correct, but I'm not sure.
 
tnich said:
Edit: I do not think your second answer is correct - that answer ignores the force of friction on box.

Ohh, I get it now, thank you for your help!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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