How to Calculate Work Done from an Acceleration vs. Time Graph?

[CAllensworth]

Homework Statement

Given Data: Graph, 2 kg particle.I am asked to calculate the work done from 0 to 4 meters, or some other interval. Normally I would simply multiply the FORCE of the given particle by multiplying 6m/s^2 times 2kg to create a new force vs. displacement graph, and then calculate the area under the linear piecewise curve. I am doing something wrong

Homework Equations

[/B]

The Attempt at a Solution

Calculate the forces as 2kg*acceleration to get the variable force y-values. Calculate area under curve... does not work according to book. Do not seem have enough info for an integral solution.

Sample attempt for Work (J) done from 1 to 2 meters..

1/2(1m*(2kg*18 m/s^2))

 

[Simon Bridge]

You appear to have an acceleration vs displacement graph (bottom horizontal axis).
Work is the area under the force vs displacement graph.
From 1-2m the area is a triangle ... which you did.
What is the problem?

 

[NTW]

In the first meter, judging from the picture, the triangle under the line has an area of (1 m*18 m/s²)/2 = 9 m²/s²

Now you multiply that by the 2kg of mass, and get 18 Joule. The units are right: acceleration [LT^-2] by displacement [L] by mass [M], hence ML²T^-2.

That way, you can continue the integration 'by eye'...

 

[CAllensworth]

I found out that I misreading the graph from the book. The book labels the scale a_s = 6m/s². Every horizontal line represented was actually 2m/s², with the graph maxing out at 6m/s². Unfortunately, a_s was actually the MAX value for the graph. My problem was with reading the scale incorrectly. The teacher was kind enough to go over the problem with me where I realized that each increment of the y-value was 2 m/s², instead of 6 m/s².

Note: apologies if I mislabeled the last graph. Please let me know if this is wrong.

Corrected Graph Accel vs Displacement Graph, this is a better representation of what was in the book:

While a_s = 6m/s

Solution:

a) Read the graph carefully, and locate all notations, even if you require a microscope.
b) Convert y-values on the graph from acceleration to FORCE via F=ma:

6m/s²(2 kg)=12 Newtons, hold the figs

i. Re-graph as a graph of F (x), and calculate the area under the curve for the W.

ii. For finding the velocity at x meters, take the integral of F(x)...

Antiderivative of F(x) results in 1/2mv²
InteGRAL over a an interVAL.
(I'm new to integration)

W=W=1/2mv² - 1/2mv²₀

Which is coincidental:

ΔKE=KE-KE₀

And where initial velocity is zero.

v=√[(2*J)/(m)] → v=√[(2*J)/(2 kg)]
​

Does the particle change direction? Answer: what has the greatest energy? The energy for a slowdown or the energy for a speedup?

This allows for solving Work Done, and Velocity from a(x). Acceleration as it relates to Displacement.​

 

[Simon Bridge]

ii. For finding the velocity at x meters, take the integral of F(x)...

... or use the work-energy theorem.