Graphical Work and Impulse 1 Shown below is a graph of the 1-dimensional, net fo

Question

Graphical Work and Impulse 1 Shown below is a graph of the 1-dimensional, net force as a function of position, F(x), that you exert on a particle of mass 20 kg. dynamically generated plot

Note 1: It may be easier to solve these problems in a different order than they are asked.

Note 2: F = 0 at x = 14.100 m.

The particle starts at x = -1 m with a velocity of 4.800 m/s and reaches x = 18.00 m under the action of the force in the graph.

What is the impulse that you give to the particle as it moves from x = -1.00 m until it reaches x = 18.00 m.

Remember that impulse is a vector, so in 1-dimension it can be either positive or negative. 1pts Submit Answer Tries 0/15

For the situation described above, what is the work done on the particle by the force from x = -1.00 m to x = 18.00 m. 1pts Submit Answer Tries 0/15

For the situation described above, what is the velocity of the particle when it reaches x = 18.00 m. 1pts Submit Answer Tries 0/15

For the situation described above, what is the impulse that the particle gives to you as it moves from x = -1.00 m until it reaches x = 18.00 m. Remember that impulse is a vector, so in 1-dimension it can be either positive or negative. 1pts Submit Answer Tries 0/15 Post Discussion

Explanation / Answer

B) Workdone = area under F-x curve

= (5 - (-1))*(-7) + 0.5*(14.1 - 5)*(-7) + 0.5*(18-14.1)*3

= -68 J

C) Apply Work-energy throrem

Wnet = change in kinetic energy

Wnet = (1/2)*m*(v2^2 - v1^2)

v2^2 - v1^2 = 2*Wnet/m

V2^2 = v1^2 + 2*Wnnet/m

v2 = sqrt(v1^2 + 2*wnet/m)

= sqrt(4.8^2 + 2*(-68)/20 )

= 4.03 m/s

A) Impulse on the particle by you = change in momentum

= m*(v2 - v1)

= 20*(4.03 - 4.8)

= -15.4 N.s or kg.m/s

D) Impulse on you by particle = +15.4 N.s or kg.m/s

Related Questions

Navigate

• Browse (All)
• Browse G
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.