Net Force of object changing velocity

In summary, the problem involves finding the magnitude of the unbalanced force acting on a supertanker that travels 3.5km and reaches a speed of 4.1km/h from rest. The solution involves using the equation Fnet = ma and converting the velocities into m/s and the distance into m. After some calculations, the answer is 1.9 * 10^4.
  • #1
ital_dj
31
0
[SOLVED] Net Force of object changing velocity

Hi, I'm studying for a test, so there might be a few of these in the next hour or so.

Homework Statement


A supertanker of mass 1.0 X 10[tex]^{8}[/tex] kg travels 3.5km, reaching a speed of 4.1km/h from rest. What was the magnitude of the unbalanced force acting on it?

Homework Equations


Fnet = ma
v[tex]^{2}_{2}[/tex] = v[tex]^{2}_{1}[/tex] + 2a[tex]\Delta[/tex]d

The Attempt at a Solution


I tried (4.1km/h)[tex]^2[/tex] = 2a(3.5km)
[tex]\frac{16.81km/h^2}{3.5km}[/tex] = 2a
4.80km = 2a
2.4km/h = a

and then I tried converting it by dividing it by 60 twice, and sliding to the right 3 decimal places, then inserting it into the Fnet = ma.

I'm really stuck, and that's not good if I have a unit test tomorrow :frown:
 
Last edited:
Physics news on Phys.org
  • #2
When using F=ma, you need m to be in kg and a to be in m/s^2
So, convert the velocities into m/s and the distance into m
 
  • #3
I'll try it again but instead beforehand.. I did it after, though, does that make a difference?
EDIT: still not working out though.
 
Last edited:
  • #4
ital_dj said:
I'll try it again but instead beforehand.. I did it after, though, does that make a difference?

Not really but it seems correct that [itex]a=2.4kmh^{-2}[/itex]. Just multiply by 1000 and divide by 3600 and it should be correct.
 
Last edited:
  • #5
The answer in the book is 1.9 * 10[tex]^4[/tex]

By multiplying 2.4 times 1000 and dividing by 3600 gives me 0.67. Still not correct. :S
 
  • #6
ital_dj said:
The answer in the book is 1.9 * 10[tex]^4[/tex]

By multiplying 2.4 times 1000 and dividing by 3600 gives me 0.67. Still not correct. :S

Divide by 3600^2 . Sorry,told you to do the wrong thing.
 
  • #7
OK Thanks.
 

1. What is the net force of an object?

The net force of an object is the overall force acting on the object, taking into account all the individual forces that are acting on it. It is a vector quantity, meaning it has both magnitude and direction.

2. How is net force related to an object's change in velocity?

According to Newton's second law of motion, the net force acting on an object is directly proportional to its acceleration, which in turn is directly proportional to its change in velocity. This means that the greater the net force acting on an object, the greater its change in velocity will be.

3. What factors affect the net force of an object?

The net force of an object is affected by the magnitude and direction of individual forces acting on it, as well as the mass of the object. The direction of the net force is determined by the vector sum of all the individual forces.

4. How do you calculate the net force of an object?

To calculate the net force of an object, you need to add up all the individual forces acting on it, taking into account their direction. If the forces are acting in the same direction, you can simply add them together. If they are acting in opposite directions, you need to subtract the smaller force from the larger one.

5. Can the net force of an object be zero?

Yes, the net force of an object can be zero if all the individual forces acting on it cancel each other out. This means that the object's acceleration will be zero, and its velocity will remain constant. Objects at rest or moving at a constant velocity have a net force of zero.

Similar threads

  • Introductory Physics Homework Help
Replies
10
Views
307
  • Introductory Physics Homework Help
Replies
6
Views
861
  • Introductory Physics Homework Help
Replies
6
Views
967
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
13
Views
2K
  • Introductory Physics Homework Help
3
Replies
97
Views
11K
  • Introductory Physics Homework Help
Replies
10
Views
6K
  • Introductory Physics Homework Help
Replies
5
Views
1K
Replies
8
Views
6K
  • Introductory Physics Homework Help
Replies
9
Views
2K
Back
Top