Box Mechanics: Solving for Speed, Energy, Force, and Work

In summary, the student pulls a 100 kg box along the x-axis and its position x changes according to the equation x=.5tcubed+2t. The speed at t=0 is 2.
  • #1
dbb2112
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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


honestly, I have no idea.
 
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  • #2
ok what i have so far is that the velocity = 1.5t^2+2, which means the speed at 0= 2.
Then when it asks for the kinetic energy I am not sure if it would be 1/2(100)(2^2) or 1/2 (100)(1.5t^2+2).
 
  • #3
anyone
 
  • #4
Hi dbb2112,

dbb2112 said:
ok what i have so far is that the velocity = 1.5t^2+2, which means the speed at 0= 2.
Then when it asks for the kinetic energy I am not sure if it would be 1/2(100)(2^2) or 1/2 (100)(1.5t^2+2).

Your answer for part a looks right to me.

For part b, neither one is quite right. The first form 1/2(100)(2^2) is the kinetic energy at t=0.

The second one:

1/2 (100)(1.5t^2+2)

is closer, but you need to square the speed. What do you get for the other answers?
 
  • #5
O you i just forgot to put the squared. But for the net force F=ma so i would get 100(3t) or 300t for the net force?
 
  • #6
dbb2112 said:
O you i just forgot to put the squared. But for the net force F=ma so i would get 100(3t) or 300t for the net force?

That looks right to me.
 
  • #7
Um about the kinetic energy again whenever i have 1/2(100)(1.5t^2+2)^2 would i have to square the 1.5 t part or could i leave it as 50(1.5t^2+2)^2
 
  • #8
dbb2112 said:
Um about the kinetic energy again whenever i have 1/2(100)(1.5t^2+2)^2 would i have to square the 1.5 t part or could i leave it as 50(1.5t^2+2)^2

I believe this last expression is right (I'm not sure what you mean by square the 1.5 t part).
 
  • #9
nevermind. About the power being delivered I am not really sure what to do here.
 
  • #10
dbb2112 said:
nevermind. About the power being delivered I am not really sure what to do here.

What two ways of calculating power are there; what are their formulas and which one applies to this case?
 
  • #11
Power=work/time?
 
  • #12
dbb2112 said:
Power=work/time?


No, that is the average power (over an interval). What's the other?
 
  • #13
power= force*velocity
 
  • #14
dbb2112 said:
power= force*velocity

That's right.

How would you find the net work done from t=0 to t=2?
 
  • #15
honestly, i am not sure.
 
  • #16
dbb2112 said:
honestly, i am not sure.

What is the work equal to? In other words, why is it impotant? For example, if I tell you that 10J of work is done on an object, what do you know about that object?
 
  • #18
How about the change in Kinetic energy is equal to the net work
 
  • #19
I know i but when you got offline i wanted someone else and therefore posted a new one.
 
  • #20
dbb2112 said:
How about the change in Kinetic energy is equal to the net work

dbb2112 said:
I know i but when you got offline i wanted someone else and therefore posted a new one.

That's correct about the net work.

However, duplicate threads are not allowed on this forum. Please stay with the original thread for this problem.
 
  • #21
o ok thanks though
 

FAQ: Box Mechanics: Solving for Speed, Energy, Force, and Work

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