# How much work, speed and resistive force

1. Oct 13, 2015

### Haveagoodday

1. The problem statement, all variables and given/known data

A crane is slowly lifting a large box of mass 2 kg by means of a thick (but

massless) rope, from the ground to a height of 10.0 m.

a) How much work does the crane do on the box? How much work does

gravity do on the box?

b) The rope suddenly breaks and the box falls to the ground. What is its

speed as it reaches the ground?

c) Sofar, we have assumed that there is no friction of any kind. Now consider

the case when there is a resistive (air drag) force on the box as it drops down

from an initial height of 10,0 m. The resistive force is modelled by

with a = 0,2kg/m. What is the terminal speed vT of the box? Using Newton’s

second law, write down the differential equation for v(t).

d) The solution to this differential equation is of the form:

where tanh is the hyperbolic tangent function (look it up), and C is some

integration constant. If t = 0 is the time when the rope breaks, what should

C be? By direct integration, find x(t); the distance fallen as a function of

time. You will need that

where cosh is the hyperbolic cosine function (look it up). Find by insertion,

whether 1.67 s, 6.71 s or 7.16 s is (approximately) the time it takes do drop

to the ground.

3. The attempt at a solution
a) W=196 J
b) v=14 m/s
c) vT=96m/s and the differential equation --> v(t)= sqrt(vT(1-Ce^(-gt/(2v/vT))))
d) C=arctanh 0= 0
x(t)=sqrt(a/mg)*v*t
x(7.16)=10.13 m

2. Oct 13, 2015

### billy_joule

Do you have a question?

3. Oct 13, 2015

### Haveagoodday

well, sort of. I have calculated each problem, but i am not sure if the results are right?

4. Oct 14, 2015

### haruspex

In a), there are two questions, but you only provided one answer.
In c), I don't understand how you got such a large terminal velocity from the given data. Maybe you wrote out the question details wrongly.
In c), you are asked for the differential equation. You appear to have answered with the solution to the equation instead.
In d), you are asked for x as a function of t, but your answer also has a variable v in it. (I have not checked whether your answer is a correct statement.)

5. Oct 14, 2015

### hefalomp

How did you get 96m/s in c, i believe that is wrong, and how did you find the diff eq. for v(t)

6. Oct 14, 2015

### jensjensen

The differential equation is
dv/dt=g+a/m*v^2. If you solve it you will get the terminal velocity 9.90 m/s, so i think it is right

7. Oct 14, 2015

### jensjensen

You can find terminal velocity putting mg-av^2=0 and solve for v. In b you got 14 without friction. Since we have friction, the terminal velocity has to be less then 14 m/s

8. Oct 14, 2015

### Hesch

Well, I've calculated the fall numerically. Here are the results (attached):

So I find that x(1.6728s ) = 10.00213m

#### Attached Files:

• ###### Fall.txt
File size:
51.9 KB
Views:
114
9. Oct 14, 2015

### haruspex

That doesn't follow. The 14 was not a terminal velocity, it was still accelerating. With very small drag, the terminal velocity coild have been more.
That said, I agree with your answer. (It must be quite a large box for its mass.)

10. Oct 14, 2015

### jensjensen

On a), wouldn't the work be 0 since you do it in a 90 degree. W=F*s*cos90=0

Last edited: Oct 14, 2015
11. Oct 14, 2015

### haruspex

I don't understand. Where are you getting an angle of 90 degrees from. Angle between what force and what displacement?

12. Oct 14, 2015

### jensjensen

My bad. The angle between the force and the displacement is of course 0 degree.