Point particle model for a person pushing a block on ice

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Homework Statement
The following image shows a person pushing a large block on a horizontal ice surface in a straight line to the right at a constant speed.

The figure presents a person pushing a large block toward the right. The block is slightly larger than the person. The person and the block are both on a horizontal surface labeled Ice.

The mass of the block is 20 kg. Frictional forces between the block and the ice are negligible. The block has a wide cross-sectional area, so air resistance is acting on the block, but air resistance on the person is negligible. The person's shoes do not slip on the ice.

As the person pushes the block, the person moves with the same constant speed as the block. In one situation, the person pushes the block with a force of 78 N moving it at a constant speed of 4.00 m/s
(I attached the image with the problem below)

a) Using a dot as a particle model (representing the person), draw and label the forces (not components) exerted on the person. Each force must be represented by a distinct arrow starting on, and pointing away from, the dot.
b) The person now stops and releases the block. Determine the magnitude and direction of the block's acceleration at the instant the block is released. Show your work.
Relevant Equations
F = ma
a) For the model, I'm not sure how to include all the forces I think are present. Gravity, normal force from the ice, contact force from ice to person, and person to ice. Is there also a normal force keeping the person's hands from going "into" the ice?

b) As for part b, I would think to use newtons second law, but I'm not sure if that's correct because the force of the person is no longer acting on the block
78 = 20a
a= 78/20
a=3.9m/s^2?

image given with question:
Screen Shot 2023-09-30 at 12.12.46 PM.png
my attempt at a model part a:
Screen Shot 2023-09-30 at 12.16.26 PM.png
 
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  • #2
Note that it's better to say point particle, rather than dot particle.
 
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  • #3
salqmander said:
Homework Statement: The following image shows a person pushing a large block on a horizontal ice surface in a straight line to the right at a constant speed.

The figure presents a person pushing a large block toward the right. The block is slightly larger than the person. The person and the block are both on a horizontal surface labeled Ice.

The mass of the block is 20 kg. Frictional forces between the block and the ice are negligible. The block has a wide cross-sectional area, so air resistance is acting on the block, but air resistance on the person is negligible. The person's shoes do not slip on the ice.

As the person pushes the block, the person moves with the same constant speed as the block. In one situation, the person pushes the block with a force of 78 N moving it at a constant speed of 4.00 m/s
(I attached the image with the problem below)

a) Using a dot as a particle model (representing the person), draw and label the forces (not components) exerted on the person. Each force must be represented by a distinct arrow starting on, and pointing away from, the dot.
b) The person now stops and releases the block. Determine the magnitude and direction of the block's acceleration at the instant the block is released. Show your work.
Relevant Equations: F = ma

a) For the model, I'm not sure how to include all the forces I think are present. Gravity, normal force from the ice, contact force from ice to person, and person to ice. Is there also a normal force keeping the person's hands from going "into" the ice?

b) As for part b, I would think to use newtons second law, but I'm not sure if that's correct because the force of the person is no longer acting on the block
78 = 20a
a= 78/20
a=3.9m/s^2?

image given with question:
View attachment 332813my attempt at a model part a:
View attachment 332815
There's only one contact force acting on the person from the block. That's ##F'_c## in your diagram. The block isn't pulling the person forward, so ##F_c## is the force exerted by the person on the block.
 
  • #4
PeroK said:
Note that it's better to say point particle, rather than dot particle.
👍
 
  • #5
PeroK said:
There's only one contact force acting on the person from the block. That's ##F'_c## in your diagram. The block isn't pulling the person forward, so ##F_c## is the force exerted by the person on the block.
thank you

is part b correct?
 
  • #6
salqmander said:
is part b correct?
No. You need to list all the forces on the block.
 
  • #7
PeroK said:
No. You need to list all the forces on the block
There was a force of friction which is negligible because the ice is not slippery so are there no more forces acting on the ice? will it continue at a speed of 4m/s so 0 acceleration?
 
  • #8
PeroK said:
There's only one contact force acting on the person from the block. That's ##F'_c## in your diagram. The block isn't pulling the person forward, so ##F_c## is the force exerted by the person on the block.
What about friction from the ground?
 
  • #9
haruspex said:
What about friction from the ground?
I think because the surface is ice friction is negligible, right?
 
  • #10
salqmander said:
I think because the surface is ice friction is negligible, right?
"The person's shoes do not slip on the ice."
 
  • #11
haruspex said:
"The person's shoes do not slip on the ice."
yea i'm not sure
 
  • #12
haruspex said:
"The person's shoes do not slip on the ice."
do you know how to solve part b? i'm not sure what to do my textbook is not very clear
 
  • #13
salqmander said:
yea i'm not sure
If the only horizontal force on the person is the 78N reaction force from the ice, why isn't the person accelerating away from the ice?
 
  • #14
salqmander said:
F = ma
a = F/m
a = g x m/ m
a=g
a= 9.8m/s^2?
This makes no sense.
What are the horizontal forces on the ice while the person is pushing? What equation relates them?
What forces change immediately the person stops pushing?
 
  • #15
yes i realized that g shouldn't be in a horizontal question, let me see
 
  • #16
salqmander said:
do you know how to solve part b? i'm not sure what to do my textbook is not very clear
You have not solved part a yet. What are the horizontal forces on the person while pushing the ice? What equation relates them?
 
  • #17
to the person, i believe there is friction force from the block,
f = \mu N
 
  • #18
salqmander said:
to the person, i believe there is friction force from the block,
f = \mu N
The person is pushing the block horizontally, not (we assume) up or down. So why would there be friction between the person and the block?
 
  • #19
haruspex said:
The person is pushing the block horizontally, not (we assume) up or down. So why would there be friction between the person and the block?
oh that makes sense. is it just contact force?
 
  • #20
salqmander said:
oh that makes sense. is it just contact force?
Between person and block, yes. The person is pushing on the block with a force of 78N, so what does Newton say about the force of the block on the person?
 
  • #21
78N back to the person
 
  • #22
salqmander said:
78N back to the person
Right. But the person is not accelerating away from the block, so what other horizontal force acts on the person?
 
  • #23
haruspex said:
Right. But the person is not accelerating away from the block, so what other horizontal force acts on the person?
is it static friction
 
  • #24
haruspex said:
What about friction from the ground?
We never got that far! I was expecting the OP to realise that an unopposed horizontal force ##F_c## couldn't be correct.

salqmander said:
oh that makes sense. is it just contact force?
You need to have a serious think about what Newton's law tell you:

1) The first law implies that unless an object is accelerating, there can be no net force on it.

2) The second law tells you how to calculate the acceleration from the mass and the net external force.

3) The third law tells you that forces come in pairs. Those pairs, however, are exerted on different objects. Third laws pairs never apply to the same object. That's why your ##F_c## was wrong.

I'll leave the thread to @haruspex, but I think you haven't really understood how all three of Newton's laws apply to this scenario.

Personally, I would start by listing all the forces on both the man and the block. Even if the question doesn't ask for it, there's no reason not to do it.
 
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  • #25
salqmander said:
is it static friction between the persons feet and the surface
Yes. And what is the magnitude of that?
 
  • #26
anyone has the asnwer yet lol?
 
  • #27
LynnS said:
anyone has the asnwer yet lol?
The OP either gave up, or solved it without updating us.
 

FAQ: Point particle model for a person pushing a block on ice

What is the point particle model?

The point particle model is a simplification in physics where an object is treated as if all its mass is concentrated at a single point. This model helps in analyzing the motion and forces acting on the object without considering its size, shape, or internal structure.

How does the point particle model apply to a person pushing a block on ice?

In the context of a person pushing a block on ice, both the person and the block can be treated as point particles. This means we consider the forces and motion without worrying about their actual dimensions. The focus is on the net force exerted by the person on the block and the resulting acceleration of the block.

What forces are involved when a person pushes a block on ice?

Several forces are involved: the applied force by the person on the block, the frictional force between the block and the ice (which is usually very small due to the low friction of ice), the gravitational force acting on both the person and the block, and the normal force exerted by the ice surface on the block.

How does friction affect the motion of the block on ice?

Friction plays a crucial role in determining the motion of the block. On ice, friction is significantly reduced, which means the block can slide more easily when pushed. The low frictional force means that even a small applied force can result in noticeable acceleration of the block.

Can the point particle model account for rotational effects?

No, the point particle model does not account for rotational effects. It simplifies the analysis by ignoring any rotational motion or torque that might be present. If rotational effects are significant, a more detailed model that considers the object's shape and distribution of mass would be necessary.

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