Work done by a varying force of a block

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SUMMARY

The discussion centers on calculating the work done on a 5.00 kg block moving along the x-axis under a varying net force, with an initial speed of 2.0 m/s at x=0. The work done when the block is displaced from x=0 to x=8 is determined to be 25 Joules, calculated by finding the area of a trapezoid and subtracting the area of a triangle. Using the work-energy theorem, the final speed at x=8 is calculated to be 3 m/s, confirming the accuracy of the calculations presented.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with the work-energy theorem
  • Ability to calculate areas of geometric shapes (trapezoids and triangles)
  • Basic knowledge of kinematics and dynamics
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  • Review the work-energy theorem in detail
  • Practice calculating work done by varying forces
  • Explore kinematic equations for motion with variable acceleration
  • Learn about graphical analysis of force vs. displacement relationships
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Homework Statement


A block with mass 5.00kg, moves along the x-axis with zero friction. The block experiences a net force Fx that varies with it's position x. When x=0, v (initial speed) is equal to 2.0 m/s

Find the work done on the block when it is displaced from x=0 to x=8

What is the final speed at x=8

This is not the original problem statement ... I had to translate it to English in order to post it here. :p

Se attached image for the graph of the relationship between the force and displacement.

Homework Equations





The Attempt at a Solution


Calculated the area in the trapezoid, and subtracted the area of the triangle. 30J-5J=25J, I think this is right answer?

Then I used the relation between work and kinetic energy

W=delta K=(1/2)m*v_f^2 - (1/2)m*v_i^2
where the subscript i and f stands for initial and final speed

Solved this for v_f and got the answer 3 m/s

Are these answers correct? Unfortunately, I don't have the answers, since this is a problem I got from my physics professor, not from the textbook.

Thanks in advance. :)
 

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Yes, you calculated the area under that curve correctly and applied the work energy theorem (work=change in kinetic energy) to calculate the final velocity at x=8. Everything seems fine to me.
 

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