Free Fall With Air Resistance Problem

[hardygirl989]

Homework Statement

A small piece of Styrofoam packing material is dropped from a height of 2.00 m above the ground. Until it reaches terminal speed, the magnitude of its acceleration is given by a = -g + bv. After falling 0.500 m, the Styrofoam effectively reaches terminal speed, and then takes 5.00 s more to reach the ground. (a) What is the value of the constant b? (b) What is the acceleration at t = 0? (c) What is the accleration when the speed is 0.150 m/s?

Homework Equations

ƩF=ma
W = weight
FD = resistant force

The Attempt at a Solution

I drew a free body diagram with the resistance force pointing up and the weight pointing down, where up is positive.

ƩF=W-FD=ma
-mg + bv = ma
-(1)g + bv = (1)a
-g + bv = a
bv-g = 0
b = g/v

I am stuck here because I am not sure how to find the terminal velocity. Do I do v=x/t=0.05/0.150=10/3≈3.33 m/s? That would make b = 9.8/(10/3)=2.94Kg/s, which seems too high...Can anyone help? Thanks.

 

[ehild]

A small piece of Styrofoam packing material is dropped from a height of 2.00 m above the ground... After falling 0.500 m, the Styrofoam effectively reaches terminal speed, and then takes 5.00 s more to reach the ground.

What distance does the Styrofarm travel with the terminal speed?

ehild

 

[Chestermiller]

Homework Statement

I am stuck here because I am not sure how to find the terminal velocity. Do I do v=x/t=0.05/0.150=10/3≈3.33 m/s? That would make b = 9.8/(10/3)=2.94Kg/s, which seems too high...Can anyone help? Thanks.

The terminal speed is when the acceleration is zero.

 

[R Morrison]

During the first 0.5 meter fall, the velocity increases from 0 m/s to the terminal velocity. After the object falls 0.5 meter, the velocity remains constant.

The velocity of the object as it falls the last 1.5 meters is the terminal velocity.
Distance = terminal velocity * time
1.5 = terminal velocity * 5
Terminal velocity = 1.5 ÷ 5 = 0.3 m/s

This is the velocity of the object, after falling the first 0.5 meter.

a = -9.8 + bv
At t = 0, v = 0
a = -9.8

At terminal velocity, acceleration= 0
-9.8 + bv = 0
-9.8 = b * v
b = -9.8 ÷ v
v = terminal velocity
b = 9.8 ÷ 0.3
b = 32⅔ = 32.667

Now that we know the value of b, we can write the equation for acceleration.

a = -9.8 + 32.667 * v

(c) What is the accleration when the speed is 0.150 m/s?
Terminal velocity = 0.3 m/s
So this speed occurs during the time, when the object was accelerating/
Use the equation for acceleration!
a = -9.8 + 32.667 * 0.150
a = -4.9 m/s^2

 

[Nickie]

To find the terminal velocity, we can use the formula v = v0 + at, where v0 is the initial velocity and a is the acceleration. At the point where the Styrofoam reaches terminal speed, the acceleration is 0 since it is no longer accelerating. Thus, we can set v = v0 + at = 0, and solve for v0.

Using this value of v0, we can then calculate the value of b using the equation b = (a + g)/v0. This will give us a more accurate value for b, rather than using the value of v calculated from the given data.

To find the acceleration at t = 0, we can use the equation a = -g + bv, where v is the initial velocity. Since the Styrofoam is dropped from rest, the initial velocity is 0, and thus the acceleration at t = 0 is simply -g.

To find the acceleration when the speed is 0.150 m/s, we can use the same equation a = -g + bv, but this time we know the value of v. Thus, we can plug in the values of g, b, and v and solve for a.

Overall, the key to solving this problem is understanding the concept of terminal velocity and using the appropriate equations to find the values of b and a at different points in time.