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A 2kg particle moving along the x-axis experiences the net force shown at right. The particle’s velocity is 3.0m/s at x = 0m.
A) At what position is the particle moving the fastest?
B) What is the particle’s velocity at x=8m? Show your work

A 2kg particle moving along the xaxis experiences the net force shown at right The particles velocity is 30ms at x 0m A At what position is the particle moving class=

Respuesta :

a. Fastest speed at x >= 8m. The fastest ACCELERATION is at x=6m
b. Speed is 5 m/s

Initial velocity is 3.0 m/s
Initial energy is 9 Joules.
Area under curve is 16 Joules
Energy at 8m is 9+16 = 25 joules
And for a particle that masses 2 kg to have 25 joules,
it needs a velocity of 5 m/s

Remember, area of triangle is 1/2 base * height. And the curve above is a triangle with base = 8 meters, height = 4 Newtons.

And Kinetic energy is E = 1/2 M V^2

Kalahira
A. Particle is moving fastest at x = 8m B. Particle's velocity at x=8m is 5.0 m/s A. The force the particle experiences ramps up from 0 to 4 Newtons, then ramps down to 0 Newtons. Since no frictional forces are mentioned, all of the forces applied to the particle accelerates it. So its maximum velocity is reached after 8 meters and it will continue on at that velocity forever. B. We need to calculate the area under the graph which since it's a triangle is 1/2 base times height. The base is 8 meters and the height is 4 Newtons. So the total area is 0.5 * 8m * 4N = 16 Nm = 16 kb m^2/s^2 Now the particle's initial velocity is 3.0 m/s, and it's initial kinetic energy is 0.5 M V^2, so E = 0.5 2kg (3m/s)^2 E = 1 kg * 9 m^2/s^2 E = 9 kg m^2/s^2 Given the 16 Nm added, the energy at x=8m will be 9 Nm + 16 Nm = 25 Nm And using the formula for kinetic energy, substitute the known values, and solve for V. 25 Nm = 0.5 * 2 kg * V^2 25 Nm = 1 kg * V^2 25 kg m^2/s^2 = 1 kg * V^2 25 m^2/s^2 = V^2 5 m/s = V

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