How to solve for the work done by a force on an object in the x direction?

  • Thread starter Thread starter chocolatelover
  • Start date Start date
  • Tags Tags
    Work
AI Thread Summary
To calculate the work done by the force F(2xi + 5yi)N on an object moving in the x direction from the origin to x=4.96m, the integral of the force must be evaluated. The correct approach involves recognizing that there is no displacement in the y direction, leading to a focus solely on the x component. The integral simplifies to W = integral from 0 to 4.96 of (2x)dx, resulting in a total work of 41.6J. Several participants noted errors in the initial calculations and emphasized the importance of correctly applying the dot product in this context. Ultimately, the discussion highlights the need for careful evaluation of integrals in physics problems.
chocolatelover
Messages
238
Reaction score
0
"work" problem

Homework Statement


A force F(2xi+5yi)N acts on an object as it moves in the x direction from the origin to x=4.96m. Find the work W=integral vector F(dr) done by the force on the object


Homework Equations





The Attempt at a Solution



integral 0 to 4.96 (2xi+5yi)N=2x^2/2|0 to 4.96=24.6

integral 5yi=5y^2/2=5(4.96)/2=61.5

61.5+24.6=
86.1

Could someone please tell me if this is correct?

Thank you very much
 
Physics news on Phys.org
Incorrect. There's no displacement in the y, that is, j direction.

Judging by your other posts on related topics, I would recommend another thorough reading up on this chapter.
 
Would it be like this?

Integral (2xi+5yj)(idx)|0 to 4.96=
integral 0 to 4.96(2x)dx=
2xx|0 to 4.56=
2x^2|0 to 4.56=
41.6

Work done in the y direction=0

Total work=41.6J

Thank you
 
Integral of 2x is not 2x^2. Might want to recheck that.
 
chocolatelover said:
Would it be like this?

Integral (2xi+5yj)(idx)|0 to 4.96=

It's a dot product: (2xi+5yj).(idx)

integral 0 to 4.96(2x)dx=
2xx|0 to 4.56=
2x^2|0 to 4.56=
41.6

(Why did it become 4.56?)

Antiderivative of 2x is 2x^2/2.

Correct your calculation.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
View full post »

Thread 'Struggling to understand the concept of this question (ice cube in water optics question)'

TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
View full post »

Similar threads

How do I find the work done for a 100 N downward force?
Replies
4
Views
2K
Resistance force changing the velocity of an object
Replies
8
Views
1K
Work done by a force down a ramp
Replies
11
Views
2K
Solving a Motion Problem with Work-Energy Theorem
Replies
23
Views
2K
Spring constant formlula for Force and Work Done
Replies
12
Views
2K
My Mistake? Understanding Friction Force & Work Done on Snowy & Icy Surfaces
Replies
29
Views
2K
Work done by non-conservative force while particle moves in circle
Replies
25
Views
2K
Calculating the work done from an equation for variable force
Replies
2
Views
2K
Work done pushing a spring from the side
Replies
9
Views
2K
Tension Force Problem (applying Work-Kinetic Energy Theorem)
Replies
16
Views
1K

Hot Threads

Recent Insights

Back
Top