Questions: Item Trial 1 Empty Cart Trial 2 Empty Cart Trial 3 With 200g Trial 4 With 200g --- --- --- --- --- Total Mass of Cart (kg) .2482 .2482 .4482 .4482 Work Done (Area) (N-m) .096 .091 .096 .217 Velocity, Initial (m/s) 0 0 0 0 Velocity, Maximum (m/s) .701 .679 .808 1.040 Change in Kinetic Energy (J) .0609 0.0572 .1463 .46617

Item  Trial 1 Empty Cart  Trial 2 Empty Cart  Trial 3 With 200g  Trial 4 With 200g
---  ---  ---  ---  ---
Total Mass of Cart (kg)  .2482  .2482  .4482  .4482
Work Done (Area) (N-m)  .096  .091  .096  .217
Velocity, Initial (m/s)  0  0  0  0
Velocity, Maximum (m/s)  .701  .679  .808  1.040
Change in Kinetic Energy (J)  .0609  0.0572  .1463  .46617
Transcript text: Item | Trial 1 Empty Cart | Trial 2 Empty Cart | Trial 3 With 200g | Trial 4 With 200g --- | --- | --- | --- | --- Total Mass of Cart (kg) | .2482 | .2482 | .4482 | .4482 Work Done (Area) (N-m) | .096 | .091 | .096 | .217 Velocity, Initial (m/s) | 0 | 0 | 0 | 0 Velocity, Maximum (m/s) | .701 | .679 | .808 | 1.040 Change in Kinetic Energy (J) | .0609 | 0.0572 | .1463 | .46617
failed

Solution

failed
failed

Solution Steps

Step 1: Understanding the Data
  • The table provides data from four trials involving a cart, with and without an additional 200g mass.
  • Key variables include total mass, work done, initial and maximum velocity, and change in kinetic energy.
Step 2: Analyzing Mass and Work Done
  • Trials 1 and 2 involve an empty cart with a total mass of \(0.2482 \, \text{kg}\).
  • Trials 3 and 4 involve a cart with an additional 200g, making the total mass \(0.4482 \, \text{kg}\).
  • Work done is measured in Newton-meters (N-m) and varies across trials.
Step 3: Calculating Kinetic Energy
  • Kinetic energy change is calculated using the formula \(\Delta KE = \frac{1}{2} m (v_{\text{max}}^2 - v_{\text{initial}}^2)\).
  • Initial velocity is \(0 \, \text{m/s}\) for all trials, simplifying the formula to \(\Delta KE = \frac{1}{2} m v_{\text{max}}^2\).
  • Compare the calculated change in kinetic energy with the provided values to verify consistency.

Final Answer

\(\boxed{0.46617 \, \text{J}}\)

Was this solution helpful?
failed
Unhelpful
failed
Helpful