PHYSICS:F/SCI...W/MOD..-UPD(LL)W/ACCES
PHYSICS:F/SCI...W/MOD..-UPD(LL)W/ACCES
9th Edition
ISBN: 9780357001417
Author: SERWAY
Publisher: CENGAGE L
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Chapter 11, Problem 20P

A 5.00-kg particle starts from the origin at time zero. Its velocity as a function of time is given by

v = 6 t 2 i ^ + 2 t j ^

where v is in meters per second and t is in seconds. (a) Find its position as a function of time. (b) Describe its motion qualitatively. Find (c) its acceleration as a function of time, (d) the net force exerted on the particle as a function of time, (e) the net torque about the origin exerted on the particle as a function of time, (f) the angular momentum of the particle as a function of time, (g) the kinetic energy of the particle as a function of time, and (h) the power injected into the system of the particle as a function of time.

(a)

Expert Solution
Check Mark
To determine

The position of particle as a function of time.

Answer to Problem 20P

The position of particle as a function of time is (2t3i^+t2j^) m .

Explanation of Solution

Given information:

The mass of particle is 5.00 kg and the velocity of particle is v=6t2i^+2tj^ .

The formula to calculate position of particle is,

r=vdt

Substitute (6t2i^+2tj^) for v in above equation to find r .

r=(6t2i^+2tj^)dt=63t3i^+22t2j^=(2t3i^+t2j^) m

Conclusion:

Therefore, the position of particle as a function of time is (2t3i^+t2j^) m .

(b)

Expert Solution
Check Mark
To determine

To describe: The motion of the particle qualitatively.

Answer to Problem 20P

the particle is moving in the x-y plane turning to move more parallel to the x-axis.

Explanation of Solution

The motion of a particle is described by the change in position vector (either in magnitude or orientation) with time.

The position of the particle is (2t3i^+t2j^) in coordinate system. As the particle starts from origin at t=0 therefore, it moves in the first quadrant. The velocity of the particle gets faster and faster with the time and it is turning to the move parallel to the x-axis in the x-y plane.

Conclusion:

Therefore, the particle is moving in the x-y plane turning to move more parallel to the x-axis.

(c)

Expert Solution
Check Mark
To determine

The acceleration of the particle as a function of time.

Answer to Problem 20P

The acceleration of the particle as a function of time is (12ti^+2j^) m/s2 .

Explanation of Solution

Given information:

The mass of particle is 5.00 kg and the velocity of particle is v=6t2i^+2tj^ .

The formula to calculate acceleration of particle is,

a=dvdt

Substitute (6t2i^+2tj^) for v in above equation to find a .

a=d(6t2i^+2tj^)dt=(12ti^+2j^) m/s2

Conclusion:

Therefore, the acceleration of the particle as a function of time is (12ti^+2j^) m/s2 .

(d)

Expert Solution
Check Mark
To determine

The net force exerted on the particle as a function of time.

Answer to Problem 20P

The net force exerted on the particle as a function of time is (60ti^+10j^) N .

Explanation of Solution

Given information:

The mass of particle is 5.00 kg and the velocity of particle is v=6t2i^+2tj^ .

The formula to calculate force exerted on the particle is,

F=ma

  • m is the mass of particle.

Substitute 5.00 kg for m and (12ti^+2j^) m/s2 for a in above equation to find F .

F=(5.00 kg)((12ti^+2j^) m/s2)=(60ti^+10j^) N

Conclusion:

Therefore, the net force exerted on the particle as a function of time is (60ti^+10j^) N .

(e)

Expert Solution
Check Mark
To determine

The net torque about the origin exerted on the particle as a function of time.

Answer to Problem 20P

The net torque about the origin exerted on the particle as a function of time is 40t3k^N-m .

Explanation of Solution

Given information:

The mass of particle is 5.00 kg and the velocity of particle is v=6t2i^+2tj^ .

The formula to calculate torque exerted on the particle is,

τ=r×F

Substitute (2t3i^+t2j^) m for r and (60ti^+10j^) N for F in above equation to find τ .

τ=(2t3i^+t2j^) m×(60ti^+10j^) N=(2t3i^×60ti^)+(2t3i^×10j^)+(t2j^×60ti^)+(t2j^×10j^) N-m=(20t3k^60t3k^) N-m=40t3k^N-m

Conclusion:

Therefore, the net torque about the origin exerted on the particle as a function of time is 40t3k^N-m .

(f)

Expert Solution
Check Mark
To determine

The angular momentum of the particle as a function of time.

Answer to Problem 20P

The angular momentum of the particle as a function of time is 10t4k^kgm2/s .

Explanation of Solution

Given information:

The mass of particle is 5.00 kg and the velocity of particle is v=6t2i^+2tj^ .

The formula to calculate angular momentum of the particle is,

L=m(r×v)

Substitute 5.00 kg for m , (2t3i^+t2j^) m for r and (6t2i^+2tj^) for v in above equation to find L .

L=(5.00 kg)[(2t3i^+t2j^) m×(6t2i^+2tj^) m/s]=(5.00 kg)[(2t3i^×6t2i^)+(2t3i^×2tj^)+(t2j^×6t2i^)+(t2j^×2tj^)] kgm2/s=(20t4k^30t4k^) kgm2/s=10t4k^kgm2/s

Conclusion:

Therefore, the angular momentum of the particle as a function of time is 10t4k^kgm2/s .

(g)

Expert Solution
Check Mark
To determine

The kinetic energy of the particle as a function of time.

Answer to Problem 20P

The kinetic energy of the particle as a function of time is (90t4+10t2)J .

Explanation of Solution

Given information:

The mass of particle is 5.00 kg and the velocity of particle is v=6t2i^+2tj^ .

The formula to calculate kinetic energy of the particle is,

KE=12m(vv)

Substitute 5.00 kg for m and (6t2i^+2tj^) for v in above equation to find KE .

KE=12(5.00 kg)((6t2i^+2tj^)m/s(6t2i^+2tj^)m/s)=12(5.00kg)(36t4+4t2)m2/s2=(90t4+10t2)J

Conclusion:

Therefore, the kinetic energy of the particle as a function of time is (90t4+10t2)J .

(h)

Expert Solution
Check Mark
To determine

The power injected into the system as a function of time.

Answer to Problem 20P

The power injected into the system as a function of time is (360t3+20t)W .

Explanation of Solution

Given information:

The mass of particle is 5.00 kg and the velocity of particle is v=6t2i^+2tj^ .

The formula to calculate power of the particle is,

P=ddt(90t4+10t2)J=(360t3+20t)W

Conclusion:

Therefore, the power of the particle as a function of time is (360t3+20t)W .

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