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Shown below is a graph of the 1-dimensional force FYP,x(x), that you exert on a particle of mass 8 kg. The point 0,0 is in the lower-left corner. Each vertical grid line is 1 N, and each horizontal grid line is 1 m. The particle starts at x=4 m with a velocity of 12.19 m/s and reaches x=22 m under the action of only the force in the graph, i.e. the force shown is the net force. What is the x component of the impulse that you give to the particle as it moves from x=4 m until it reaches x=22 m? Remember that impulse is a vector, so you will need to take into account directions. In particular, you will need to determine whether or not the particle turns around. Because impulse is a vector, in 1 dimension it can be either positive or negative

Shown below is a graph of the 1-dimensional force FYP,x(x), that you exert on a particle of mass 8 kg. The point 0,0 is in the lower-left corner. Each vertical grid line is 1 N, and each horizontal grid line is 1 m. The particle starts at x=4 m with a velocity of 12.19 m/s and reaches x=22 m under the action of only the force in the graph, i.e. the force shown is the net force. What is the x component of the impulse that you give to the particle as it moves from x=4 m until it reaches x=22 m? Remember that impulse is a vector, so you will need to take into account directions. In particular, you will need to determine whether or not the particle turns around. Because impulse is a vector, in 1 dimension it can be either positive or negative

College Physics
College Physics
10th Edition
ISBN:9781285737027
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter5: Energy
Section: Chapter Questions
Problem 43P: The system shown in Figure P5.43 is used to lift an object of mass m = 76.0 kg. A constant downward...
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Shown below is a graph of the 1-dimensional force FYP,x(x), that you exert on a particle of mass 8 kg. The point 0,0 is in the lower-left corner. Each vertical grid line is 1 N, and each horizontal grid line is 1 m.

The particle starts at x=4 m with a velocity of 12.19 m/s and reaches x=22 m under the action of only the force in the graph, i.e. the force shown is the net force.

What is the x component of the impulse that you give to the particle as it moves from x=4 m until it reaches x=22 m? Remember that impulse is a vector, so you will need to take into account directions. In particular, you will need to determine whether or not the particle turns around. Because impulse is a vector, in 1 dimension it can be either positive or negative.

Shown below is a graph of the 1-dimensional force Fyp(x), that you exert on a particle of mass 8 kg. The point 0,0 is in the lower-left corner. Each vertical grid line is 1 N, and each horizontal grid line is 1 m. The particle starts at x = 4 m with a velocity of 12.19 m/s and reaches x = 22 m under the action of only the force in the graph, i.e. the force shown is the net force. Work What is the work done on the particle by the force from x = 4 m to x = 22 m? W = 170 J Particle Speed What is the speed of the particle when it reaches x = 22 m? Ux = 13.82 m/s Impulse of You on Particle What is the x component of the impulse that you give to the particle as it moves from x = 4 m until it reaches x = 22 m? Remember that impulse is a vector, so you will need to take into account directions. In particular, you will need to determine whether or not the particle turns around. Because impulse is a vector, in 1 dimension it can be either positive or negative. ym,x
Transcribed Image Text:Shown below is a graph of the 1-dimensional force Fyp(x), that you exert on a particle of mass 8 kg. The point 0,0 is in the lower-left corner. Each vertical grid line is 1 N, and each horizontal grid line is 1 m. The particle starts at x = 4 m with a velocity of 12.19 m/s and reaches x = 22 m under the action of only the force in the graph, i.e. the force shown is the net force. Work What is the work done on the particle by the force from x = 4 m to x = 22 m? W = 170 J Particle Speed What is the speed of the particle when it reaches x = 22 m? Ux = 13.82 m/s Impulse of You on Particle What is the x component of the impulse that you give to the particle as it moves from x = 4 m until it reaches x = 22 m? Remember that impulse is a vector, so you will need to take into account directions. In particular, you will need to determine whether or not the particle turns around. Because impulse is a vector, in 1 dimension it can be either positive or negative. ym,x
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