What was the speed of the object when it was released from the table?

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In summary, the conversation was about a popular pastime involving pushing an object close to the edge of a table without it falling off. The object was pushed too hard and fell off the edge, landing a certain distance from the table. The task was to calculate the object's speed as it was released, using the coefficient of kinetic friction. The solution involved finding the time it took for the object to hit the ground and the velocity in the x direction as it left the table. The final answer was 3.68 m/s, but the online homework program indicated a rounding error or incorrect use of significant figures. The correct approach is to use the difference of the velocities squared equal to 2 times acceleration times distance.
  • #1
Nightrider55
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Homework Statement



A popular pastime is to see who can push an object closest to the edge of a table without its going off. You push the 100 g object and release it 2.20 m from the table edge. Unfortunately, you push a little too hard. The object slides across, sails off the edge, falls 0.500 m to the floor, and lands 25.0 cm from the edge of the table.

If the coefficient of kinetic friction is 0.300, what was the object's speed as you released it?

Homework Equations




y = (1/2)*g*t^2 therefore t = (2*y/g)^(1/2)
x = v2*t therefore x = v2*(2*y/g)^(1/2)

The Attempt at a Solution



I worked everything out and I got 3.60 m/s for the speed when released but my online homework program says "Not quite. Check through your calculations; you may have made a rounding error or used the wrong number of significant figures." I checked my work and I am not sure where I went wrong.

Here are the values I got

V2=.7836 m/s velocity when it leaves the table
T(3-2) time it takes to hit the ground once leaving the table=.319 sec
A0=-2.94m/s^2
V0=3.5966

To find V0 I used X1=(-Vox^2)/(2Ax) then solved for Vox any insight as to where I went wrong?
 
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  • #2
I thought I'd give this problem a shot, so here it goes. First, I find the time it takes for the object to hit the ground once its off the edge of the table:

0.500 m = (1/2) (9.8 m/s^2) t^2 -> t = Sqrt(1/9.81) -> t = 3.19E-1 s

next, I solve for the velocity in the x direction as the object leaves the table:

.25 m = Vedge (3.19E-1 s) -> Vedge = .25/3.19E-1 -> Vedge = 7.83E-1 m/s

Here is where it gets fuzy, I solved for Vinitial in terms of energy:

(1/2) m Vinitial^2 - F d = 1/2 m Vedge^2 -> Vinitial = Sqrt(Vedge^2 +2 Fd/m)
with Fd/m = (mue) g d = .3*9.8*2.20

So Vinitial = Sqrt((7.83E-1)^2 + 2(6.468)) = 3.68 m/s

Assuming I'm right, your answer is close. You probably rounded off at some point. I think significant figures are only 2, due to .25 being given, so the answer would round off to 3.7? Significant figures are not my strong point. Take this with a grain of salt.
 
  • #3
Yes. You should be using v0^2-v2^2=2*a*x.
 
  • #4
bhimberg is correct. You should be taking the difference of the velocities squared equal to 2*a*x. It's not round off, it's a concept problem.
 

Related to What was the speed of the object when it was released from the table?

1. What factors affect the distance an object will travel when launched off a table?

The distance an object will travel when launched off a table is affected by several factors, including the initial velocity of the object, the angle at which it is launched, and the force of gravity. Other factors such as air resistance and the surface of the table may also play a role.

2. How does the angle of launch affect the trajectory of an object launched off a table?

The angle of launch can greatly impact the trajectory of an object launched off a table. Launching at a higher angle will result in a shorter distance traveled but a higher peak height, while launching at a lower angle will result in a longer distance traveled but a lower peak height.

3. What is the maximum height an object can reach when launched off a table?

The maximum height an object can reach when launched off a table is dependent on the initial velocity and angle of launch. In ideal conditions, the maximum height will occur at a launch angle of 90 degrees, but in reality, air resistance and other factors will affect the actual height reached.

4. How does air resistance affect the distance an object will travel when launched off a table?

Air resistance can significantly affect the distance an object will travel when launched off a table. The presence of air resistance will cause the object to slow down and reduce its overall distance traveled. This effect is more pronounced for objects with a larger surface area or a lower initial velocity.

5. Can the surface of the table affect the trajectory of an object launched off it?

The surface of the table can definitely affect the trajectory of an object launched off it. A rough or uneven surface can cause the object to bounce or change direction, while a smooth and flat surface will allow for a more predictable trajectory. Additionally, the coefficient of friction between the object and the table can also impact its trajectory.

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