Finding the Magnitude of the Force Acting on an Object

In summary: In this equation, ##v_o## is the initial velocity, and ##a## is the acceleration. So, the equation says that the acceleration is the same as the change in velocity (i.e. the rate of change of velocity).In summary, the student is trying to find the rate of change of velocity, and is not able to do so because they do not have derivatives of the equation.
  • #1
correction_tape
2
0
Homework Statement
The velocity (m/s) of an object at time (t) is given by v = -4.0 + 2.0t.

(1) Calculate the magnitude of the force F acting on the object of mass m = 4.0kg.
(2) Calculate the kinetic energy (K) of the object of mass m = 4.0kg at time t = 3.0s.
(3) Setting the initial position x(subscript 0) = 0.0m of the object at time t = 0.0s, calculate the time (t>0.0s) at the position x=21m of the object.
Relevant Equations
F = ma
a = dv/dt
So far, I've only attempted the first number:

F = (4kg) [(-4+2t)-0]/t

?

Been thinking how to cancel out the time, but I really have no idea how to proceed. Was I supposed to cancel it even? Can someone please help and tell me how you would answer this? (And maybe show a solution as well so I can make sense of what to do when I encounter these problems again)

By the way, this is not homework. I'm just studying from exam papers online.
 
Physics news on Phys.org
  • #2
Do you understand how derivatives work? You have divided velocity by t, not taken its derivative with respect to t.
 
  • #3
Orodruin said:
Do you understand how derivatives work? You have divided velocity by t, not taken its derivative with respect yo t.
Sorry, not very much. How do you do that?
 
  • #4
A derivative is a measure of how fast something changes with a variable. In this case how velocity changes with time, which by definition is acceleration. Also in this case, velocity changes linearly with time so acceleration is constant.

Derivatives is generally something you will want to study before trying to dig deeper into kinematics (or any subfield of physics really). This is an entire subject by itself so it is not really suitable for a PF post.

Are you familiar with the SUVAT equations?
 
  • #5
This can be solved without derivatives but you must know SUVAT equations and the related kinematics theory.
Also Newton's 2nd law is needed but you have that in the list of related equations in the OP.

Orodruin said:
Derivatives is generally something you will want to study before trying to dig deeper into kinematics (or any subfield of physics really)
Yes , derivative and integrals, differential and integral calculus in general is almost absolutely basic stuff (along with linear algebra) for the way physics are being taught today.
 
  • #6
correction_tape said:
Sorry, not very much. How do you do that?
In simpler language, your ##a=dv/dt## means ##a## is the slope of the graph of ##v## as a function of ##t##.
Since, in your case, this graph is straight line, the expression for ##a## is $$a= { v(t+\Delta t) - v(t)\over \Delta t} = { -4+2( t+ \Delta t ) - ( -4+2t ) \over \Delta t}= ...$$

##\ ##
 
  • Like
Likes Orodruin
  • #7
Hi @correction_tape. Welcome to PF. I’m guessing you are working at an introductory/shool level, so differentiating (i.e. using calculus) may not be appropriate.

The velocity (in m/s) of an object at time t (in s) is given by v(t) = -4.0 + 2.0t.

Look at the equation carefully. Can you tell how much v changes every time t increases by 1 second?

That’s the rate of change of v, i.e. the acceleration.
 
  • #8
correction_tape said:
v = -4.0 + 2.0t.
Compare this to the equation ##v=v_o+at##.
 
  • Like
Likes Delta2

What is the formula for finding the magnitude of the force acting on an object?

The formula for finding the magnitude of the force acting on an object is F = m x a, where F is the force in Newtons, m is the mass of the object in kilograms, and a is the acceleration in meters per second squared.

How do you determine the direction of the force acting on an object?

The direction of the force acting on an object can be determined by using a coordinate system and assigning positive and negative directions. The force will act in the direction of the acceleration, so if the acceleration is positive, the force will act in the positive direction and vice versa.

What factors affect the magnitude of the force acting on an object?

The magnitude of the force acting on an object is affected by the mass of the object and the acceleration it experiences. The greater the mass of the object or the greater the acceleration, the greater the magnitude of the force.

Can the magnitude of the force acting on an object be negative?

Yes, the magnitude of the force acting on an object can be negative. This would indicate that the force is acting in the opposite direction of the acceleration, as negative acceleration would result in a negative force.

How is the magnitude of the force acting on an object related to its motion?

The magnitude of the force acting on an object is directly related to its motion. The greater the force acting on an object, the greater the acceleration and therefore the greater the change in motion of the object.

Similar threads

  • Introductory Physics Homework Help
Replies
15
Views
413
  • Introductory Physics Homework Help
Replies
2
Views
970
  • Introductory Physics Homework Help
Replies
6
Views
490
  • Introductory Physics Homework Help
Replies
17
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
30
Views
2K
  • Introductory Physics Homework Help
Replies
17
Views
590
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
2
Replies
62
Views
3K
  • Introductory Physics Homework Help
Replies
8
Views
1K
Back
Top