Show the velocity graph of a 4.7 kg object as it moves along the x-axis

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SUMMARY

The discussion focuses on calculating momentum and acceleration for a 4.7 kg object moving along the x-axis using the formula F = m•a. Participants provided examples of calculating force by multiplying the mass (4.7 kg) with various velocities (6 m/s, 12 m/s, and 3 m/s), yielding momentum values of 28.2 kg m/s, 56.4 kg m/s, and 14.1 kg m/s, respectively. The conversation also highlights the relationship between velocity and momentum, emphasizing that the y-axis of a velocity-time graph represents velocity in m/s, which can be multiplied by mass to find momentum. Additionally, the equation F = dp/dt is introduced as relevant for understanding force in terms of momentum change over time.

PREREQUISITES
  • Understanding of Newton's Second Law (F = m•a)
  • Knowledge of momentum (p = m•v)
  • Familiarity with velocity-time graphs
  • Basic calculus concepts related to derivatives (dp/dt)
NEXT STEPS
  • Study the implications of Newton's Second Law in different physical scenarios
  • Explore the concept of momentum in elastic and inelastic collisions
  • Learn how to interpret and analyze velocity-time graphs
  • Investigate the relationship between force, mass, and acceleration in various contexts
USEFUL FOR

Students in physics, educators teaching mechanics, and anyone interested in understanding the principles of motion and force calculations.

Cb8
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Homework Statement
Figure 1 shows the velocity graph of a 4.7 kg object as it moves along the x-axis.
A) What is the x- component of the net force acting on this object at t = 1s?
B) At 3s?
C) At 7s?
Relevant Equations
F=M•a
If force= m•a would the answer just be the y-axis • 4.7 at the specific time such as
A) 4.7 • 6 = 28.2?
B) 4.7 • 12= 56.4
C) 4.7 • 3 = 14.1
 
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Post Figure 1. (If you have a velocity-time graph, how would you find the acceleration at any point?)
 
I solved it! Sorry for the late response, if you’re still wondering.
image.jpg
 
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Cb8 said:
If force= m•a would the answer just be the y-axis • 4.7 at the specific time such as
A) 4.7 • 6 = 28.2?
B) 4.7 • 12= 56.4
C) 4.7 • 3 = 14.1
Since the y-axis shows velocity (in m/s) multiplying the y-axis by the mass of 4.7 (in kg) yields momentum (in kg m/s).

A relevant equation would be: ##F=\frac{dp}{dt}##
 

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