Mechanics of a Sliding Box: Calculating Kinetic Energy, Net Force, and Work Done

AI Thread Summary
The discussion revolves around calculating various physics parameters for a 100 kg box being pulled along a rough surface, with its position described by the equation x = 0.5t^3 + 2t. Participants explore how to determine the box's speed at t=0, its kinetic energy, net force, and power as functions of time. The importance of using derivatives to find velocity and acceleration is emphasized, with kinetic energy calculated as 200 J at t=0. The conversation also highlights the need to integrate to find work done over a specified interval and suggests that the work done can be determined by the change in kinetic energy between two time points.
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Homework Statement



A 100 kg box is being pulled along the x-axis by a student. The box slides along the a rough surface and its position x varies accoring to the equation x=.5tcubed +2t, where x is in meters and t is in seconds.

a. Determine the speed of the box at t=0
b. Determine the following as functions of time t.
i. The kinetic energy of the box
ii. the net force acting on the box
iii. The power being delivered to the box.
c. Calculate the net workdone on the box in interval t=0 t t=2
d. Indicate below wheter the work done on the box by the student in the interval t=0 to t=2 wold e greater than, less than or eequal to the answer in part c.

Homework Equations





The Attempt at a Solution

 
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dbb2112 said:

Homework Statement



A 100 kg box is being pulled along the x-axis by a student. The box slides along the a rough surface and its position x varies accoring to the equation x=.5tcubed +2t, where x is in meters and t is in seconds.

a. Determine the speed of the box at t=0
b. Determine the following as functions of time t.
i. The kinetic energy of the box
ii. the net force acting on the box
iii. The power being delivered to the box.
c. Calculate the net workdone on the box in interval t=0 t t=2
d. Indicate below wheter the work done on the box by the student in the interval t=0 to t=2 wold e greater than, less than or eequal to the answer in part c.

How would you think to approach the problem?

You have a displacement/time graph ... so how do you read velocity?
 
I don't really know how to approach the probelm or how i read velocity
 
before i go there this is just guessing but would i take the derivitive of the equation to find the velocity?
 
dbb2112 said:
before i go there this is just guessing but would i take the derivitive of the equation to find the velocity?

Since that's what velocity is - the rate of change of position wrt time - then yes that's what you will need to be doing.
 
ok, so at t=0 the velocity would be 2m/s.
It then asks to Determine the following as functions of time t. I do not understand what that means.
 
i. The kinetic energy of the box
ii. the net force acting on the box
iii. The power being delivered to the box.

You have the velocity function.
You will need the acceleration function too.

KE - 1/2*m*v2 = 1/2*m*vt2

F = m*a

P = w*/t = F*d/t = Ft*vt

Basically you are using these functions in place of the usual functions you might employ in determining more traditional uniform gravitational kinematics.
 
i do not really understand what finding those as functions of time means still. But i am assuming that i should just plug in the velocity i found for kinetic energy and mass they already give you to obtain the answer of 200J. And then derive the velocity again to get the acceleration of 3t? that is where I kind of get lost.
 
  • #10
dbb2112 said:
i do not really understand what finding those as functions of time means still. But i am assuming that i should just plug in the velocity i found for kinetic energy and mass they already give you to obtain the answer of 200J. And then derive the velocity again to get the acceleration of 3t? that is where I kind of get lost.

In b) they are asking you for these things as functions of time. (Remember a is the derivative of v and equals 3*t, which is what you apparently have determined.)

In c) you need to integrate the product of F and X from t=0 to t=2.
 
  • #11
so with the acceleration being 3t would i sub anything in there or just multiply 3 by the mass.
 
  • #12
dbb2112 said:
so with the acceleration being 3t would i sub anything in there or just multiply 3 by the mass.

No. Wherever you need to know a is where you would sub in 3*t, just as for velocity wherever you have v you sub in (3*t2/2 +2) wherever you have v.

These variables are functions of time. And the equations in b they are asking for are in terms of these.
 
  • #13
O ok so the force woudl be 300t.
 
  • #14
and for the next part, i would integrate 300t *.5tcubed+ 300t*2t?
 
  • #15
dbb2112 said:
and for the next part, i would integrate 300t *.5tcubed+ 300t*2t?

That looks correct. It would be the definite integral over the time given.
 
  • #16
the only problem now is I am not really sure how to do that.
 
  • #17
would it be 30t^5+300t^3? and then just plug in the times they give me?
 
  • #18
dbb2112 said:
would it be 30t^5+300t^3? and then just plug in the times they give me?

It's 1/3 of 600 isn't it?
 
  • #19
o yes so it would be 30t^5+200t^3 but after that I am not too sure what to do would i plug in 0 and then plug in 2
 
  • #20
could you tell me why we integrated there also?
 
  • #21
dbb2112 said:
could you tell me why we integrated there also?

That comes from the definition of Work, being the integral of the product of Force in the direction of displacement and displacement.

Since they are both functions of time and you are required to identify the amount of work over that time then you need to integrate over the time to get the correct summation of all the incremental Δ FxD's

See also:
http://en.wikipedia.org/wiki/Mechanical_work#Force_and_displacement
 
  • #22
im still not grasping it, could you demonstrate what you mean please?
 
  • #23
dbb2112 said:
im still not grasping it, could you demonstrate what you mean please?

Read the Wikipedia link.
 
  • #24
instead of integrating, could i just find both kinetic energies and subtract them?
 
  • #25
dbb2112 said:
instead of integrating, could i just find both kinetic energies and subtract them?

Does that yield the same answer?
 
  • #26
i don't know i still can not integrate, but my guess would be yes
 
  • #27
ok thank you for all your help but its almost 1 oclock and i have school tomorrow so could you leave a post for me of how i would answer the last part of my question so that i could look at it in the morning. Again, thank you for your help.
 
  • #28
dbb2112 said:
i don't know i still can not integrate, but my guess would be yes

Actually since the question is about the work on the block that will be simply given by the change in KE. KE at t=2 less the KE at t=0 should do it.
 

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