The graph below shows the net force on a particle in the x-direction as a function of its position along the x-axis. The mass of the particle is m = 3kg. Ę (N) 4202+ -2 -4 Part 1 x (m) 1 2 3 If the particle has a velocity of v = 2m/s when x = 0m, what is the particle's speed when x = 2.5m? v = number (rtol=0.05, atol=1e-08) m/s ? Part 2 At what value of x does the particle have its maximum kinetic energy? x = number (rtol=0.05, atol=1e-08) m ? Part 3 What is the particle's maximum kinetic energy? K = number (rtol=0.05, atol=1e-08) J
The graph below shows the net force on a particle in the x-direction as a function of its position along the x-axis. The mass of the particle is m = 3kg. Ę (N) 4202+ -2 -4 Part 1 x (m) 1 2 3 If the particle has a velocity of v = 2m/s when x = 0m, what is the particle's speed when x = 2.5m? v = number (rtol=0.05, atol=1e-08) m/s ? Part 2 At what value of x does the particle have its maximum kinetic energy? x = number (rtol=0.05, atol=1e-08) m ? Part 3 What is the particle's maximum kinetic energy? K = number (rtol=0.05, atol=1e-08) J
Principles of Physics: A Calculus-Based Text
5th Edition
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter7: Conservation Of Energy
Section: Chapter Questions
Problem 2OQ: An athlete jumping vertically on a trampoline leaves the surface with a velocity of 8.5 m/s upward....
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Transcribed Image Text:The graph below shows the net force on a particle in the x-direction as a function of its
position along the x-axis. The mass of the particle is m = 3kg.
Ę (N)
4202+
-2
-4
Part 1
x (m)
1
2
3
If the particle has a velocity of v = 2m/s when x = 0m, what is the particle's
speed when x = 2.5m?
v = number (rtol=0.05, atol=1e-08)
m/s
?
Part 2
At what value of x does the particle have its maximum kinetic energy?
x = number (rtol=0.05, atol=1e-08)
m
?
Part 3
What is the particle's maximum kinetic energy?
K =
number (rtol=0.05, atol=1e-08)
J
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