Work Problem w/variable force- Did I do it right?

In summary: Part b)The work done is$$W=\int_0^{x_{\max }} {{F_x}\,dx} = \int_0^{x_{\max }} {F_0 \,{\rm{sin}}\left( {cx} \right)\,dx} = \left. { - \frac{{{F_0}}}{c} \,{\rm{cos}}\left( {cx} \right)} \right|_0^{x_{\max }} = \frac{{{F_0}}}{c}\left( {1 - {\rm{cos}}\left( {c{x_{\max }}} \right)} \
  • #1
bcjochim07
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Homework Statement


A particle of mass m starts from xo= 0m with vo> 0 m/s. The particle experiences the variable force Fx= Fosin(cx) as it moves to right along the x-axis, where Fo and c are constants.

a)At what position xmax does the force first reach a maximum value?

b) What is the particle's velocity as it reaches xmax? Give your answer in terms of m, vo, Fo, & c.

Homework Equations





The Attempt at a Solution



Here is my idea for part a:

Take the derivative and set it equal to 0
Fx'=Fo*c*cos(cx)
0=Fo*c*cos(cx)
cos(cx)=0
cx= pi/2
xmax=pi/2c Is this correct?

Then for part b:

I took the integral of Fosin(cx):

(Fo/c) * (-cos(pi/2) + 1) = Fo/c This is the work done.

Then using the work-kinetic energy theorem:

.5mvo^2 + Fo/c = .5 mv1^2 where v1 is the velocity at x max

.5vo^2 +Fo/c*m = .5v1^2
vo^2 + 2Fo/c*m = v1^2
sqrt (vo^2 + 2Fo/c*m) = v1
 
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  • #2
It is a bit hard to follow your work, but everything seem to be correct.
 
  • #3
A particle of mass m starts from x_0=0\,{\rm m} with v_0>0\,{\rm m/s}. The particle experiences the variable force F_x =F_0 \,{\rm{sin}}\left( {cx} \right) as it moves to the right along the x-axis, where F_0 and c are constants.


the vast majority of students trying to complete this problem on mastering physics will just copy and paste this... so I'll put it here...

the answer above is completely right (TIP, be careful with the parentheses)
 

1. What is a work problem with variable force?

A work problem with variable force is a physics problem that involves calculating the work done on an object when the force applied to it varies with distance or position.

2. How do I solve a work problem with variable force?

To solve a work problem with variable force, you will need to use the formula W = ∫ F(x) dx, where W represents work, F(x) represents the varying force, and dx represents a small change in distance. You will also need to determine the limits of integration based on the given distance or position values.

3. What units are used for work and force in a variable force problem?

Work is typically measured in joules (J), while force is measured in newtons (N). However, in some cases, different units may be used, such as foot-pounds (ft-lb) for work and pounds (lb) for force. It is important to pay attention to the given units and convert them if necessary to ensure accurate calculations.

4. What are some common mistakes to avoid when solving a work problem with variable force?

Some common mistakes to avoid when solving a work problem with variable force include using the wrong formula, not paying attention to units, and not properly determining the limits of integration. It is also important to double check your calculations and make sure they make sense in the context of the problem.

5. Can I use the work-energy theorem in a work problem with variable force?

Yes, the work-energy theorem (W = ΔKE) can be used in a work problem with variable force, as long as the force is conservative. In this case, the change in kinetic energy of the object will be equal to the work done by the varying force on the object.

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