# Hi, help me to check my answers, dynamics, energy and work

## Homework Statement

Whats is the final speed of the second box(mass = 1kg) when the first one (mass = 5 kg) has descended a distance equal to 0.6 meters in the rough ramp inclined 60° respect the x axis. Consider the spring is compressed a length x = 0.2 meters. The second box is tied to the spring and the system is released from its stationary state. For this case, consider the rope, the spring and the pulley are ideal (have no mass and the rope doesn't get elonged or contracted), the dynamic friction coefficent is 0.2, and the elasticity constant of the spring k=200 Newtons/meters?
• Second box(mass = 1kg)
• First box(mass = 5 kg)
• Distance descended to 0.6 meters (rough ramp inclined 60° respect the x axis)
• The spring is compressed a length x = 0.2 meters.
• The rope, the spring and the pulley are ideal (have no mass and the rope doesn't get elonged or contracted).
• Dynamic friction coefficent is 0.2
• Elasticity constant of the spring k=200 Newtons/meters

It's the 27th excersice ## Homework Equations

• Free-body diagrams
• Work (W=Fxd)
• Mechanic Energy Balance (conservative and non conservative system)
• Cinematic equation for final speed V^2Final = V^2initial +2 a(Xfinal - Xinitial)
Being
V^2Final the square power of the speed at a t time
V^2initial the square power of the speed at a initial time
a the acceleration
(Xfinal the displacement at a the t time
Xinitial the displacement at a the initial time

## The Attempt at a Solution

The speed I calculated is equal to 3.68 m/s
The acceleration of the system when released (thanks to the free-body diagrams) = 11,29m/s^2

In order to find the speed of the second box, I first found the height where it will arrive after the spring releases, using energy balances. Then, after findig the acceleration via Free-body diagram used cinematic equations for speed according to de enunciated distance.

#### Attachments

ZapperZ
Staff Emeritus
I bet that this has an accompanying figure. Look at it from our point of view. Without the figure, do you think the problem and scenario is clear based just on what you have written?

Zz.

I bet that this has an accompanying figure. Look at it from our point of view. Without the figure, do you think the problem and scenario is clear based just on what you have written?

Zz.
Done.

haruspex
Homework Helper
Gold Member

## Homework Statement

Whats is the final speed of the second box(mass = 1kg) when the first one (mass = 5 kg) has descended a distance equal to 0.6 meters in the rough ramp inclined 60° respect the x axis. Consider the spring is compressed a length x = 0.2 meters. The second box is tied to the spring and the system is released from its stationary state. For this case, consider the rope, the spring and the pulley are ideal (have no mass and the rope doesn't get elonged or contracted), the dynamic friction coefficent is 0.2, and the elasticity constant of the spring k=200 Newtons/meters?
• Second box(mass = 1kg)
• First box(mass = 5 kg)
• Distance descended to 0.6 meters (rough ramp inclined 60° respect the x axis)
• The spring is compressed a length x = 0.2 meters.
• The rope, the spring and the pulley are ideal (have no mass and the rope doesn't get elonged or contracted).
• Dynamic friction coefficent is 0.2
• Elasticity constant of the spring k=200 Newtons/meters

It's the 27th excersice
View attachment 228331

## Homework Equations

• Free-body diagrams
• Work (W=Fxd)
• Mechanic Energy Balance (conservative and non conservative system)
• Cinematic equation for final speed V^2Final = V^2initial +2 a(Xfinal - Xinitial)
Being
V^2Final the square power of the speed at a t time
V^2initial the square power of the speed at a initial time
a the acceleration
(Xfinal the displacement at a the t time
Xinitial the displacement at a the initial time

## The Attempt at a Solution

The speed I calculated is equal to 3.68 m/s
The acceleration of the system when released (thanks to the free-body diagrams) = 11,29m/s^2

In order to find the speed of the second box, I first found the height where it will arrive after the spring releases, using energy balances. Then, after findig the acceleration via Free-body diagram used cinematic equations for speed according to de enunciated distance.
Maybe I am making some silly error, but it seems to me the box will not descend that far.
Some of the details (in English) are not entirely clear. Is the spring already compressed by 0.2m before the system is released? Is the 0.6m descent vertical discplacement or parallel to the plane (I assumed it parallel to the plane, leading to total spring compression 0.8m).
Very rough analysis: GPE lost by 5kg mass < 5g(0.6)0.87<26J.
Work done on spring =½200(0.82-0.22)=60J

Please either correct my assumptions or post detailed working (preferably both).

Maybe I am making some silly error, but it seems to me the box will not descend that far.
Some of the details (in English) are not entirely clear. Is the spring already compressed by 0.2m before the system is released? Is the 0.6m descent vertical discplacement or parallel to the plane (I assumed it parallel to the plane, leading to total spring compression 0.8m).
Very rough analysis: GPE lost by 5kg mass < 5g(0.6)0.87<26J.
Work done on spring =½200(0.82-0.22)=60J

Please either correct my assumptions or post detailed working (preferably both).

The sexond box is compressing the spring before the system is released.

The first box descends o,6 meters parallel to the plane.

But the descending of the first box (the left one) means the rising of the second one (the one tied directly to the spring).

haruspex
Homework Helper
Gold Member
The sexond box is compressing the spring before the system is released.

The first box descends o,6 meters parallel to the plane.

But the descending of the first box (the left one) means the rising of the second one (the one tied directly to the spring).
Doh! I transcribed the diagram wrongly!
Makes sense now, but I get a rather smaller number. Please post your working.

Doh! I transcribed the diagram wrongly!
Makes sense now, but I get a rather smaller number. Please post your working.

I redid it. Now what I've done is calculate what heigt the 2nd box will reach (gave equals to 0.15 meters (the height obtained was the same using 2 different reference systems).

Then calculated de acceleration of the system via free-body diagrams (Acceleration= 11,29m/s^2)

And finally decomposed the 0,6 meters of the ramp to find the y component (height) to know that the second box raises 0,52 meters when the system moves

Then used cinematic equations to calculate the final speed when the box raised (final speed= 2,89m/s)

Last edited:
haruspex
Homework Helper
Gold Member
what heigt the 2nd box will reach (gave equals to 0.15 meters
If it only rises 0.15m there is no way way the other box will move 0.6m.
calculated de acceleration
Since we do not care about time, acceleration is uninteresting. You can do it all by energy.

I still get a smaller value for the speed. Please post your working!

I redid it. Now what I've done is calculate what heigt the 2nd box will reach (gave equals to 0.15 meters

Then calculated de acceleration of the system via free-body diagrams (Acceleration= 11,29m/s^2)

And finally decomposed the 0,6 meters of the ramp to find the y component (height) to know that the second box raises 0,52 meters when the system moves

Then used cinematic equations to calculate the final speed when the box raised (final speed= 2,89m/s)
If it only rises 0.15m there is no way way the other box will move 0.6m.

Since we do not care about time, acceleration is uninteresting. You can do it all by energy.

I still get a smaller value for the speed. Please post your working!
I mean, the 0,15 meters is what the 2nd box raises by the action of the spring and the 0.52 is what the 2nd box raises by releasing the system. other value for velocity I calculated by making an energy balance from the height -0,2m (before releasing the system to the higest ponit the second box will reach is 0.89 m/s but I'm not sure about that result).

haruspex