Simple work and variable position question

Click For Summary
SUMMARY

The discussion focuses on calculating the work done by a variable force acting on a particle with mass 6.26 kg, where the position is defined by the equation x = 0.484t^3 - 33.6t. The user correctly identifies the need to integrate the force over the displacement to find work, resulting in W = -2253.8 J after evaluating the integral from 0 to 1.49 seconds. The user also notes the absence of gravitational force in the problem statement, emphasizing the need to derive force from the position function.

PREREQUISITES
  • Understanding of Newton's second law (F = ma)
  • Knowledge of calculus, specifically integration
  • Familiarity with the concept of work (W = F * s)
  • Basic physics principles regarding forces and motion
NEXT STEPS
  • Study the derivation of force from position functions in physics
  • Learn about integrating variable forces in one-dimensional motion
  • Explore the implications of non-constant forces on work calculations
  • Review the relationship between mass, acceleration, and work in physics
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and work-energy principles, as well as educators looking for examples of variable force problems.

iJamJL
Messages
58
Reaction score
0

Homework Statement


A one-dimensional force acts on a particle of mass m = 6.26 kg in such a way that its position is given by:

x = 0.484t^3 - 33.6t

Find W, the work done by this force during the first 1.49 s.


Homework Equations


W=F*s
W=mgs
Integration


The Attempt at a Solution


I just wanted to know whether I'm solving this correctly because I only have one more chance to input an answer in my online homework system (as those of you who have been helping me probably know by now. :frown:

Anyhow, this is what I've done.

W=Fs
F=mg
W=mgs
**Because the position is variable but the mass and gravity are constant, we integrate the formula given, and we get:

W=mg*∫x.dx= (.121t^4) - 16.8(t^2), from 0s to 1.49s

We come to:

W=mg*(-36.7)
W=6.26*9.81*(-36.7)= (-2253.8) J

Did I do this properly?
 
Physics news on Phys.org
The problem does not say that the particle is subjected to gravity. F=ma, and the acceleration is second derivative of position. Find the force from the given x(t) function.

ehild
 

Similar threads

Not sure which formula to use Deals with work and position
  • · Replies 8 ·
Replies
8
Views
3K
Calculating the work done from an equation for variable force
  • · Replies 2 ·
Replies
2
Views
2K
Determine the work done by the force due to gravity
  • · Replies 14 ·
Replies
14
Views
3K
Calculation of work done by a variable force
  • · Replies 8 ·
Replies
8
Views
4K
How far does the block slide? (work, spring, incline)
  • · Replies 4 ·
Replies
4
Views
2K
Rotational Mechanics Question - A Rotating Bar
  • · Replies 9 ·
Replies
9
Views
2K
How Much Power Does an Escalator Need to Operate Fully Laden?
  • · Replies 8 ·
Replies
8
Views
8K
Work done given force as a function of position
  • · Replies 2 ·
Replies
2
Views
2K
Energy/Work Problem: Friction of sliding down pole
  • · Replies 5 ·
Replies
5
Views
4K
What is the Work Done by Several Forces?
  • · Replies 11 ·
Replies
11
Views
2K