Solve Dynamics Question: 3.0L Bottle Thrown Upward with 55N Force

[jen333]

Hey!
hope someone can help me on this question:

A 3.0L (3.0kg) bottle of water is thrown vertically upward with a force of 55N. How high will it go if the force is applied for a distance of 80cm starting from rest?
This answer given is 69cm

I know that the acceleration of the bottle while the upward force is applied is 8.5m/s^2, and the applied force is 55N, and drew an FBD
Other than that, i don't know where to start to carry out this question!

 

[quasar987]

Look at the equations of motion at constant acceleration.

v(t) = v_0 + at

x(t) = x_0 + vt +0.5at²

 

[jen333]

Look at the equations of motion at constant acceleration.

v(t) = v_0 + at

x(t) = x_0 + vt +0.5at²

sorry, but i still don't understand
in those formulas, an amount of time is required isn't it?

 

[quasar987]

First of all, I would like to edit an error in my first post: The equations are

v(t) = v_0 + at

x(t) = x_0 + v_0t +0.5at²


Those are the equations that tell you how the speed and the position of a particle evolve in time when it is subject to a constamt acceleration a.

You know it starts from rest (v_0 = 0). You can also set your coordinate axes such that x_0 = 0 too. Additionally, F=ma ==> a = F/m = 55/3. Now the equations become considerably simpler:

v(t) = (55/3)t

x(t) = 0.5(55/3)t²

From those, can you find at which time t the bottle is at position x = 0.80 m ?

 

[marky]

EDIT: Forgot to include work done by gravity.

You have a force acting over a distance. That means work is done on the system.

Work done, [tex] U = Fd - mgh = (55 N)*(.8m) - (0.8m)(9.81)(3) = 20.5 J [/tex]

Work and Energy Equation [tex]T_i + U = T_f [/tex]

T is kinetic energy [tex] T = \frac{1}{2}mv^2 [/tex]

[tex] T_i = 0 [/tex] because you start from rest,[tex] v = 0[/tex], so [tex]T = 0[/tex]

[tex]T_f[/tex] will equal the work done. [tex]T_f = 20.5[/tex] J

From that you can get speed. [tex]T_f = \frac{1}{2}mv^2 = 20.5[/tex]

You have mass m, solve for velocity v. [tex]v = 3.69[/tex] m/s

Now it becomes a kinematics problem. Initial velocity is 3.69 m/s.
Final velocity is 0 (corresponding to its max height it will reach). Acceleration is -9.81 m/s^2 (gravity)

Using[tex] v_f^2 = v_i^2 + 2ad [/tex]
[tex]0 = 3.69^2 + 2(-9.81)(d) [/tex]
Solve for d
d = 0.69 m or 69 cm

 

[jen333]

thx for your help! but i have just one more question.
The acceleration that was used was -9.81mm/s^2, however, i was just wondering about the acceleration, 8.5m/s^2, of the bottle while the upward force is applied. does that have any affect on this problem? can it be used to solve the problem another way? just wondering

 

[marky]

thx for your help! but i have just one more question.
The acceleration that was used was -9.81mm/s^2, however, i was just wondering about the acceleration, 8.5m/s^2, of the bottle while the upward force is applied. does that have any affect on this problem? can it be used to solve the problem another way? just wondering

Yes, sure.
Applying [tex]v_f^2 = v_i^2 + 2ad[/tex]
Let's look at the point from x = 0 to x=0.8. This is where the applied force is being exerted.
Your FBD should have the force of gravity and the applied force. Summing F = ma you get a = 8.5 m/s^2.

[tex]v_i = 0[/tex]
[tex]v_f = \sqrt{ 2 (8.5)(0.8)} = 3.688 m/s[/tex]

The moment you let go of the bottle and let it fly upward, you have a velocity of 3.688 m/s.
Now the applied force is no longer there when you let go of the bottle (throwers hand applies the force I assume), only the force of gravity acts on the bottle.
Applying [tex]v_f^2 = v_i^2 + 2as[/tex]
[tex] 0^2 = 3.688^2 + 2(-9.81)(s)[/tex]
s = .69m

 

[marky]

Sorry if you read my original first solution. It was wrong. I forgot to include the work done by gravity. Anyways, I made the changes. There are two approaches. Both of which have the same answer. The second seems shorter. But often using work and energy can simplify things which is why I attempted that first.