Calculating Work done by the Earth and the work done by the Spring

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SUMMARY

The discussion focuses on calculating the work done by the Earth and the spring on a mass of 0.091 kg that is thrown downward from a vertical spring. For part (a), the work done by the Earth is calculated using the formula \( W = F \times d \), resulting in -0.080262 Joules, indicating that the force of gravity acts in the opposite direction to the displacement. In part (b), the work done by the spring requires calculating the change in kinetic energy as the mass slows down to 2.13 m/s after moving 0.09 m downward.

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  • Understanding of Newton's laws of motion
  • Familiarity with the work-energy principle
  • Knowledge of gravitational force calculations
  • Basic proficiency in kinetic energy formulas
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  • Calculate the work done by the spring using the change in kinetic energy formula
  • Review the concept of net work and its implications on motion
  • Explore the relationship between force, displacement, and work in different contexts
  • Study the impact of varying mass and speed on work calculations
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Paul36
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Homework Statement


A mass of 0.091 kg hangs from a vertical spring in the lab room. You pull down on the mass and throw it vertically downward. The speed of the mass just after leaving your hand is 3.80 m/s.

Homework Equations


(a) While the mass moves downward a distance of 0.09 m, how much work was done on the mass by the Earth? Include the appropriate sign.
Work done by Earth = ...Joules

(b) At the instant in part (a) when the mass has moved downward a distance of 0.09 m, the speed of the mass has decreased to 2.13 m/s. How much work was done on the mass by the spring? Include the appropriate sign.
Work done by spring = ...Joules

The Attempt at a Solution


To calculate Work I'm using the formula F times displacement
So for (a) I calculated:
0.09m X -9.8m/s/s X 0.091kg = -0.080262J
is that correct?
For part B I have no idea what to do?
Thanks if you can help.
 
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Part B mainly requires for you to calculate the difference in the kinetic energy.
 
Paul36 said:
To calculate Work I'm using the formula F times displacement
So for (a) I calculated:
0.09m X -9.8m/s/s X 0.091kg = -0.080262J
is that correct?
Except for the sign. What determines the sign of the work done is whether the force and displacement are in the same direction or not. If they are, the work done is positive.
For part B I have no idea what to do?
What's the net work done on the mass?
 

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