How Do You Calculate Force from Acceleration and Change in Direction?

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To calculate the force acting on a 5.0 kg mass with a velocity change from 2.5 m/s east to 17.5 m/s west over 3 seconds, it's essential to first determine the average acceleration, which considers direction. The change in velocity is calculated as the final velocity (17.5 m/s west) minus the initial velocity (2.5 m/s east), resulting in a total change of 20 m/s when accounting for direction. Therefore, the average acceleration is 20 m/s divided by 3 seconds, yielding approximately 6.67 m/s². Using the formula F=ma, the force is then calculated as 5.0 kg multiplied by 6.67 m/s², resulting in a force of about 33.35 N directed west. Understanding the significance of direction in vector quantities is crucial for accurate calculations in physics.
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Homework Statement



Calculate the force acting on a 5.0 kg mass which accelerates for 3. 0 s. During this time its velocity changes from 2.5 m s^-1 east to 17.5 m s^-1 west.

Homework Equations



F=ma

The Attempt at a Solution



Not sure what the acceleration is, i am only accustomed to getting a specific value e.g. 2.5 ms -1, how do i work out the acceleration, do i do:

17.5 - 2.5 = 15.0 ms west then

5.0 kg/15.0 ms west.

P.S. I started preliminary physics 1 week ago, and this is one homework question i cannot do, Thanks in advance
 
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Hello RobinsoKo

What's the definition of acceleration? (it will be average acceleration in this case).
 
The time rate of change of velocity.

Final velocity - Initial Velocity over 3 seconds?
 
RobinsoKo said:
The time rate of change of velocity.

Final velocity - Initial Velocity over 3 seconds?

Yeah, but you need to pay attention to the "east" and "west" bits of the problem. Velocity is a vector--it has both a magnitude AND direction. You can't ignore the direction! To help you out, say the mass is moving on the x-axis. Then think of east as the +x direction and west as the -x direction (or vice versa). So what do you think the average acceleration is then?
 
I worked out the acceleration to be 15ms^-1 over 3 = 5 ms-1?

So therefore the force is 25n west?
 
Not quite. You didn't take into account the direction. Remember, it's the CHANGE in velocity over 3 seconds. If you were going 2.5 m/s to the east and ended up going 17.5 m/s to the west you actually switched directions, right?

What you did was found the acceleration if you were going 2.5 m/s to the west and accelerated to 17.5 m/s to the west. So there was no directional change. You're missing the change from 2.5 m/s to the east to 2.5 m/s to the west!

You have to take the direction into account! Can you figure out how?
 
I have no idea, and this is first week into physics i am hopeless :(
 
Let's just look at a separate problem to sort of guide you there:

Say your initial velocity is 2 m/s to the west and your final velocity is 2 m/s to the east. You accelerate from your initial velocity to your final velocity in 1 second. What's your acceleration?

Let me give you an idea of what's happening.

1. First you're going 2 m/s to the west. Let this arrow represent that: <--
2. After a bit (0.5 seconds to be exact) you're going 0 m/s. Why? Because you're changing directions from the west to the east! So in the transition from going to the west to going to the east, you'll have to go 0 m/s at *some* point, right? It's like turning around in your car.
3. Now you're going 2 m/s to the east, after 1 second. Let this arrow represent that: -->

Can you tell me what your acceleration was?
 
2ms^2 east
 
  • #10
RobinsoKo said:
2ms^2 east

No. What are you using for your change in velocity?
 
  • #11
Initial - 2m/s east
Final - 2m/s west

Wait.. Is it zero?
 
  • #12
RobinsoKo said:
Initial - 2m/s east
Final - 2m/s west

Wait.. Is it zero?

No, why would it be zero? Again, you're failing to understand the relevance of direction. It's not only about magnitude. If you're driving your car down the highway at 60km/hr and you turn around, you'll have to first slow down and then speed back up to 60km/hr in the other direction. You're accelerating the whole time!

Let's set up the situation on an axis:WEST --------------------------------------------------EAST

So first you're going 2m/s to the west:

<---- 2m/s

WEST --------------------------------------------------EAST


And at the end you're going 2m/s to the east:

----> 2m/s
WEST --------------------------------------------------EAST



I don't know about you, but this looks a lot like the x-axis to me. So let's rename WEST and EAST to -x and +x:

-x <--------------------------------------------------> +x

So at first you're going 2m/s to the west. But now west is the negative x direction. So instead of having to say "2m/s to the west" each time let's just say your initial velocity is -2m/s. At the end you're going 2m/s to the east. But now the east is the positive x direction. So let's say your final velocity is +2m/s.

Now, given that acceleration is equal to

change in velocity/change in time = (final velocity - initial velocity)/change in time, tell me what your acceleration should be. Show your work.

A final, visual way to think about it is in terms of the arrows (vectors):

What arrow do I have to add to:

<----

to get

----> ? Hint: <---- + ----> = 0.

You probably haven't covered vectors yet, but you should sort of have an intuitive idea.
 
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