Particle is moved - Work-kinetic energy

In summary: You are given the initial velocity so you can determine the change in kinetic energy and from that the final kinetic energy.Finally for 3) the overall energy doesn't change - there is no energy input, the work done is all internal - so the power must be the same in both cases. So if you use the change in work in 1) to determine the change in kinetic energy you can determine the power in both cases.In summary, the conversation discusses a 2.00 kg particle being moved from (1.00, 3.00) m to (5.00, 7.00) m and the calculation of work, final velocity, and power involved. The work done by the force F =
  • #1
Fendergutt
6
0

Homework Statement



A 2.00 kg particle is moved from the coordinates (1.00, 3.00) m to (5.00, 7.00) m.

(q1) How much work is done by the force F = <2x^2, 0> on the particle? (Assume that the force is in Newton when x is in metres.)

(q2) If the particle has start velocity v_i = <0, 4.00 m/s>, what is the final velocity v_f? (Hint: the "work - kinetic energy"-theorem can be useful.)

(q3) Determine the rate of energy transfer (i.e. the effect P) from the force to the particle, at start and final position. (Answers: P_i = 0 W, P_f = 455 W.)

Homework Equations





The Attempt at a Solution



(q1) I made a position vector AB = <4, 4>. It takes the particle from (1, 3) to (5, 7). We consider the positional vector's component that is parallell with F.

--> x = 4 m --> W_F = 2 * 4^2 = 32, that is 32 N.


(q2) Start velocity v_i = <0, 4.00 m/s> is constant in y-direction since the only working force on the particle is F, but this force is perpendicular on v_i, and will therefore not alter it.

"Work - kinetic energy"-theorem:

W_total = (delta)K = (1/2)*m(v_F)^2

<--> v_F = sqrt( (2(W_F)) / m ) = sqrt( (2*32 N) / (2 kg) ) = 4sqrt(2) m/s

--> v_f = <v_F, v_i>

--> ||v_f|| = sqrt((v_F)^2 + (v_i)^2) = sqrt( (4sqrt2 m/s)^2 + (4.00 m/s)^2 ) = 4sqrt3 m/s =~ 6.9 m/s

that is in the direction of the vector AB, 45 degrees anticlockwise with the x-axis.


(q3) Effect from force on particle:

at start: P = dW/dt = F[/*v_i = F * v_i * cos(alpha) = 2*0^2*cos(pi/2) = 0.
ie 0 W

at end: P = dW/dt = F*v_f = F*v_f*cos(theta) = 2* 4^2 * 4sqrt3 * cos(pi/4) W = 64sqrt6 W =~ 157 W.

answer: at start: 0 W, at end: 455 W

I must have done something wrong here. I believe the error follows from my try at (q2). Please comment or help me with getting this problem right. Thanks a lot for your time.
 
Physics news on Phys.org
  • #2
Welcome to PF.

For 1) I would suggest that your answer isn't in the right units for work. You got 32 N and W is in N-m.

Specifically W = ∫ F(x)⋅dx

Since your Force has no component in Y then you would use x2 as your F(x). You would evaluate that over the range of x from 1 to 5. This method will yield work as N-m.

Using this result - the work supplied - and starting with an initial kinetic energy given by ½mv² after adding the work to that you can determine the velocity at (5,7)
 
  • #3

Your approach to solving this problem is correct. However, there are a few minor errors in your calculations.

For (q1), the work done by the force F is calculated correctly as 32 N. However, it is important to note that this is the magnitude of the force and not the work itself. The work is a scalar quantity and should be expressed in joules (J). Therefore, the correct answer for (q1) is 32 J.

For (q2), your approach is correct but there is a small error in your calculation for the final velocity. The correct calculation should be:

v_f = sqrt((v_F)^2 + (v_i)^2) = sqrt( (4sqrt2 m/s)^2 + (4.00 m/s)^2 ) = 4sqrt(6) m/s =~ 9.8 m/s

For (q3), your approach is also correct but there is a small error in your calculation for the rate of energy transfer at the final position. The correct calculation should be:

P_f = dW/dt = F * v_f * cos(theta) = 2 * (4sqrt6 m/s) * (4sqrt3 m/s) * cos(pi/4) = 64sqrt(18) W =~ 455 W

Therefore, the correct answers for (q1) to (q3) are:

(q1) Work done by the force F = 32 J
(q2) Final velocity v_f = 4sqrt(6) m/s =~ 9.8 m/s
(q3) Rate of energy transfer at start = 0 W, Rate of energy transfer at final position = 455 W

Overall, your approach to solving this problem is correct but there are just a few minor errors in your calculations. Keep up the good work!
 

Related to Particle is moved - Work-kinetic energy

1. What is the relationship between work and kinetic energy?

The work done on an object is equal to the change in its kinetic energy. This means that the work done on an object will either increase or decrease its kinetic energy, depending on the direction of the force applied.

2. How does the concept of work apply to the movement of particles?

The work done on a particle is equal to the force applied to the particle multiplied by the distance it moves in the direction of the force. This means that in order to move a particle, a force must be applied to it and work must be done.

3. What is the difference between work and kinetic energy?

Work and kinetic energy are related concepts, but they are not the same. Work is a measure of the force applied to an object, while kinetic energy is a measure of the object's motion. Work can change an object's kinetic energy, but they are not interchangeable.

4. Can work be negative when a particle is moved?

Yes, work can be negative when a particle is moved. This occurs when the force applied to the particle is in the opposite direction of its motion. In this case, the work done on the particle will decrease its kinetic energy.

5. How is work calculated in the context of particles and kinetic energy?

In the context of particles and kinetic energy, work is calculated by multiplying the force applied to the particle by the distance it moves in the direction of the force. This can be represented by the equation W = Fd, where W is work, F is force, and d is distance.

Similar threads

  • Introductory Physics Homework Help
2
Replies
55
Views
2K
  • Introductory Physics Homework Help
2
Replies
38
Views
2K
  • Introductory Physics Homework Help
Replies
12
Views
845
  • Introductory Physics Homework Help
Replies
9
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
414
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
14
Views
3K
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
3K
  • Introductory Physics Homework Help
Replies
3
Views
819
Back
Top