Particle is moved - Work-kinetic energy

[Fendergutt]

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.

 

[LowlyPion]

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)

 

[nlightner]

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!