Finding the Unknown Force: Acceleration and Net Force Calculation

[Jose Miranda]

Homework Statement

A 3.7 kg mass has a constant acceleration from an initial velocity of 2.33 ihat m/s to a final velocity of (18.9 ihat - 10.7 jhat) m/s in 2.71 s. Two forces act on the mass: gravity and an unknown applied force. Determine the acceleration and unknown force.[/B]

Homework Equations

The Attempt at a Solution

For the acceleration i managed to get 16.25 m/s, and for the force of gravity it is -36.26 m/s.[/B]

 

[BvU]

Hello Jose, welcome to PF

Can you show your attempt ? The numerical value of the answer isn't as important as how you found it.
And what equation would you like to use to determine the applied force ? I assume that's where you run into problems ? Where precisely ?

 

[Jose Miranda]

Hello Jose, welcome to PF

Can you show your attempt ? The numerical value of the answer isn't as important as how you found it.
And what equation would you like to use to determine the applied force ? I assume that's where you run into problems ? Where precisely ?

I believe the a I got the first time is wrong.
a = (V-Vi)/t, so a = (6.1ihat - 3.95 jhat) m/s

The problem is on on incline starting at 0,0 and having an angle below the x-line.

So the force on the x-axis is 3.7 kg times 6.1 = 22.6 N, and the y force 3.7kg time 3.95 = 14.6 N

22.6 +14.6 = 37.2 N

37.2N - 36.26 N = 0.94 N
0.94 N = unknown force

 

[lightgrav]

gravity around here, all by itself, causes acceleration 9.8 m/s² ... that's 32 ft/s² , if anybody still uses feet.
So the Force by gravity is 36.26 Newtons ... watch your units.

 

[jbriggs444]

I believe the a I got the first time is wrong.
a = (V-Vi)/t, so a = (6.1ihat - 3.95 jhat) m/s

The problem is on on incline starting at 0,0 and having an angle below the x-line.

So the force on the x-axis is 3.7 kg times 6.1 = 22.6 N, and the y force 3.7kg time 3.95 = 14.6 N

I agree with the corrected numbers. Thank you for showing your work.

22.6 +14.6 = 37.2 N

22.6 N in one direction plus 14.6 N in the direction at right angles equals what?

37.2N - 36.26 N = 0.94 N

Same problem. 37.2 N in what direction? Minus 36.26 N in what direction?

 

[Jose Miranda]

I agree with the corrected numbers. Thank you for showing your work.22.6 N in one direction plus 14.6 N in the direction at right angles equals what?Same problem. 37.2 N in what direction? Minus 36.26 N in what direction?

What stumps me is the how can i do this if there is no angle.

 

[jbriggs444]

You have referred to an "incline", an "x-axis" and a point at "0,0". Possibly there is a diagram that has not been shared? A reasonable assumption would be that gravity acts in the -y direction. You do not need an angle if you consider gravity and the y component of the unknown force together and the x component of the unknown force separately.

 

[haruspex]

What stumps me is the how can i do this if there is no angle.

The problem statement is still unclear to me.
If it is 'on an incline', there would be a normal force from the surface. Is that the unknown force, or considered a consequence of the gravity?
Please state the entire question word for word.

 

[Jose Miranda]

The problem statement is still unclear to me.
If it is 'on an incline', there would be a normal force from the surface. Is that the unknown force, or considered a consequence of the gravity?
Please state the entire question word for word.

The problem doesn't mention an incline i just assumed it was on an incline, because of its final velocity.

 

[BvU]

I believe the a I got the first time is wrong.
a = (V-Vi)/t, so a = (6.1ihat - 3.95 jhat) m/s

The problem is on on incline starting at 0,0 and having an angle below the x-line.

So the force on the x-axis is 3.7 kg times 6.1 = 22.6 N, and the y force 3.7kg time 3.95 = 14.6 N

22.6 +14.6 = 37.2 N

37.2N - 36.26 N = 0.94 N
0.94 N = unknown force

I subscribe to the a = (6.1ihat - 3.95 jhat) m/s, and also to force on the x-axis is 3.7 kg times 6.1 = 22.6 N, and the y force 3.7kg time 3.95 = 14.6 N
but from there you start losing me:

If you go one step to your right ($\hat\imath$) and one step forward ($\hat\jmath$), the result is that your displacement is √2 steps at 45 degrees.
It's called vector addition. The two forces have to be added as vectors -- if you want to add them at all. But in this exercise you don't want that. They give velocities as $\ ( a\; \hat\imath + b\; \hat\jmath\ )\$, so an answer in the same terms should be OK.​

The gravity force that works in the $\hat\jmath$ direction is $\ m\;\vec g\ = 3.7 * -9.81\; \hat\jmath\$. If the net force is $\ -14.6\; \hat\jmath\$ N (I hope you didn't forget the minus sign ?) then the applied force in the y-direction must be quite positive.

This whole escapade can take place on an inclined plane or in thin air -- if the exercise tekst says "Two forces act on the mass: gravity and an unknown applied force" then that is enough to calculate that one unknown force. Which is all the exercise wants you to determine.

 

[Jose Miranda]

I subscribe to the a = (6.1ihat - 3.95 jhat) m/s, and also to force on the x-axis is 3.7 kg times 6.1 = 22.6 N, and the y force 3.7kg time 3.95 = 14.6 N
but from there you start losing me:

If you go one step to your right ($\hat\imath$) and one step forward ($\hat\jmath$), the result is that your displacement is √2 steps at 45 degrees.
It's called vector addition. The two forces have to be added as vectors -- if you want to add them at all. But in this exercise you don't want that. They give velocities as $\ ( a\; \hat\imath + b\; \hat\jmath\ )\$, so an answer in the same terms should be OK.​

The gravity force that works in the $\hat\jmath$ direction is $\ m\;\vec g\ = 3.7 * -9.81\; \hat\jmath\$. If the net force is $\ -14.6\; \hat\jmath\$ N (I hope you didn't forget the minus sign ?) then the applied force in the y-direction must be quite positive.

This whole escapade can take place on an inclined plane or in thin air -- if the exercise tekst says "Two forces act on the mass: gravity and an unknown applied force" then that is enough to calculate that one unknown force. Which is all the exercise wants you to determine.

ok, so an answer for this can be given as vectors? (22.60ihat N -14.6 jhat N) ?

 

[BvU]

What you describe here is the net force. That isn't the answer for this exercise. I think the exercise asks for the second force.