Time to Calculate the Piano's Fall

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Homework Help Overview

The problem involves a piano being raised to the third floor of a building and subsequently falling after the rope snaps. It is set in the context of kinematics, specifically focusing on the motion of freely falling objects and the application of kinematic equations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the initial conditions of the piano's motion, including its height and initial velocity. There are questions about the appropriate kinematic equations to use and whether to treat the motion as a single segment or two separate motions. Some participants express uncertainty about the acceleration due to gravity and its implications for the initial velocity.

Discussion Status

There is an ongoing exploration of the kinematic equations relevant to the problem. Some participants have suggested using the quadratic formula to solve for time, while others have raised questions about the assumptions being made regarding initial velocity and displacement. A positive value for time has been calculated, but the discussion remains open with varying interpretations of the motion.

Contextual Notes

Participants are navigating the implications of the piano's initial upward motion and how it transitions to free fall after the rope snaps. There is a focus on understanding the definitions of initial velocity and displacement in the context of the problem.

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A piano is being raised to the third floor of a building using a rope and pulley. The piano is 6.0 m above the ground, moving upwards at 2.9 m s-1, when the rope snaps. Calculate how much time (in s) elapses before the piano hits the ground.

i need to know how to approach the problem
thanks in advance.
 
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Treat the piano like any other falling object. You are given its initial speed and height. (Hint: What's the general kinematic relationship that gives position as a function of time for uniformly accelerated motion?)
 
If we use s = ut + 1/2at^2
for the piano to reach the top
s=6
u=2.9
v=0
a=Don't know
t=?
or
s=0
u=2.9
v=?
a=What would be the a?
t=?
 
Of course you know a! Once the rope snaps, the piano is a freely falling object.

A more complete kinematic expression would be:
[tex]y = y_0 + v_0 t + 1/2 a t^2[/tex]
 
yes but then the initial velocity can't be 2.9 but will be 0 so that we could use a=g
 
is Vo = initial velocity
and Y = vertical displacement
what does Yo = ? initial displacement.
 
kingyof2thejring said:
yes but then the initial velocity can't be 2.9 but will be 0 so that we could use a=g
Why can't the initial velocity be 2.9 m/s? a = -g.

Yes, [itex]y_0[/itex] is the initial displacement.
 
if we consider these two motions
one the piano being pulled up at 2.9ms-1 but the final velocity will become
0ms-1
two the piano falling down from the 0ms-1 to final velocity with -a therefore
-displacement taking vectors upwards to be positive.

or are we considering it as one motion where the displacement is 0
initial velocity is 2.9 but then how does a = -g.
 
You could consider the motion as one continuous segment (that's how I'd do it) or break it up into two segments (up and down); done correctly, both methods will give the same answer.

At the moment the rope breaks, the piano has an initial speed (2.9 m/s) and an initial height (6.0 m). Once the rope breaks the only force acting on the piano is gravity, so a = -g. The final height is zero.

You could break the motion up into two segments (up and down), but if you know how to use that kinematic equation there is no need to do that.
 
  • #10
0 = 6 + 2.9t - 4.9t^2
how do get t?
 
  • #11
if i use the quadratic formula "and should i be using it in the first place for such basic type of problem" but anyway i get 1.441 sec as +ve value
 
  • #12
Looks good to me.
 
  • #13
thank you__________
 

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