Solving Newton's Methods: F=ma, v=at+v_0, T=ma

  • Thread starter Thread starter Effitol840
  • Start date Start date
AI Thread Summary
The discussion revolves around solving two physics problems using Newton's laws. For the freight train, the user calculates the time to accelerate from rest to 90 km/h using F=ma and v=at+v_0, but struggles with unit conversion between m/s² and km/h. In the elevator problem, the user attempts to find the tension (T) while considering the forces acting on the elevator, including gravity and tension, and is advised to establish a proper sign convention for the forces. The conversation emphasizes the importance of correctly identifying the direction of forces and applying the net force equation. Overall, the user seeks clarification on the correct application of these concepts to solve the problems accurately.
Effitol840
Messages
16
Reaction score
0
I have 2 questions:

1) A freight train has a mass of 1.1\times 10^7 kg. If the locomotive can exert a constant pull of 7.8\times 10^5 N, how long does it take to increase the speed of the train from rest to 90 km/h?

I figured I could just use the formula F=ma to get the acceleration then I plugged that into v=at+v_0 to get the time. I came up with the answer of 1267.61. I might have a problem with units because a was m/s^2 and v was in km/h. If that's the case how do I go about converting m/s^2 to km/h or vice versa?

2) An elevator of mass m moving upward has two forces acting on it: the upward force of tension in the cable and the downward force due to gravity. Let the elevator have a mass of 1200 kg and an upward acceleration of 2.1 m/s^2. Find T.

With this one I just plugged it into the equation T=ma but I don't know if this is the correct equation. I got an answer of 2520N.

If anyone could help me with this please do.
 
Physics news on Phys.org
convert your km/h in the question to m/s in this way
1km/hr = 1000m/3600secs

why did you equate T with ma? Is tension the only force causing this acceleration? Keep in mind that the DIRECTION of these forces. Assume one direction to be positive and one to be negative and write down the force equation for this. That is
ALL Net forces = Net Force = m x Net Acceleration
 
I don't know where I got that equation from. I'm still a little confused about the tension problem. I guess I don't really know how to go about solving it.
 
Effitol840 said:
I don't know where I got that equation from. I'm still a little confused about the tension problem. I guess I don't really know how to go about solving it.

Ok. Draw a box taht represents the elevator. Assume up to be positive and down to be negative. What force points up, and what force points down? Now add these forces. But use the proper sign convention as in waht points up in positive and what pints down is negative. Now which way is the elevator accelratin? Is this acceleration (using the reference system) postive or negative. What is the force due to this acceleration? Now keep in mind that
Sum of all the forces (poiting up and down) = Net Force(this may be positive or negative, depending on which way teh elevator is accelerating.
 
Ok so what I get out of that is that I'm trying to find T which is the tension on the elevator pointing up? And the force down is the gravitational force which is 9.8 m/s. The elevator is accelerating up at 2.1m/s^2. So now I need to find an equation to figure out T.
 
Last edited:
Effitol840 said:
Ok so what I get out of that is that I'm trying to find T which is the tension on the elevator pointing up? And the force down is the gravitational force which is 9.8 m/s. The elevator is accelerating up at 2.1m/s^2. So that would be T+9.8=2.1t?

What is the formula for force? Thus how do you find the force exerted by gravity? ANd the net force?

Also remember to make a REFERENCE SYSTEM. One should be positive and one be negative. Are Tensio nand gravity in the same direction? Correct the signs!
 
F=ma. Force up is possitive 2.1m/s^2 and force down is negative 9.8m/s^2. So does that mean total F=negative 7.7m/s^2?
 
Last edited:
Back
Top