This problem is related to physics, but my problem is mainly due to calculus topics so I thought I would ask here.
There is a practice problem in my mastering physics book for calculus based physics that states: "A 5.0 kg box is attached to one end of a spring with a spring constant of 80 N/m. The other end is anchored to a wall. Initially the box is at rest at the spring's equilibrium position. A rope with a constant tension of 100 N then pulls the box away from the wall. The coefficient of friction between the box and the floor is .30. How much power is being supplied by the hand or motor pulling the rope when the box has moved 50 cm?"
The solution they give has to do with writing out $K_{final} + U_{final} + \Delta E_{thermal} = K_{initial} + U_{intitial} + W_{external}$ . Then they plug in all the values known with $W_{external} = Tension * distance$ and $\Delta E_{thermal} = f_{kinetic\;friction} * distance$. They then solve for v to be 3.6 m/s. With this they give $Power = Tension * velocity = 360\;Watts$.
I understand this but I was wondering if it could be done a different way like this:
First write a Force function of distance I got $F(x) = Tension - f_{kinetic\;friction} - F_{spring} = 100N -(5kg)(9.8 \frac{m}{s^2})(.3)-(80 \frac{N}{m})(x) = -80x + 85.3$
Then I integrate with respect to position x to get $\int F(x)dx = Work = W(x) = -40x^2 + 85.3x + C$
Then I take a derivative of work with respect to time using the chain rule. $\frac {dW(x)}{dt} = Power = P(t)= -80x \frac {dx}{dt} + 85.3 \frac {dx}{dt} + 0$
So now I need to find $\frac {dx}{dt}$ which is just velocity. Going back to the force function, since $F=ma$ then $a=\frac {F}{m}$ So $a(x) = \frac {F(x)}{m} = \frac{F(x)}{5} = \frac{-80x + 85.3}{5} = -16x + 17.6$ Then integrate this to get velocity: $\int (-16x + 17.6)dx = v(x) = -8x^2 + 17.6x + c$ Since v initial is 0 I then did $v(0) = 0 = -8(0)^2 + 17.6(0) + c$ so $c = 0$. Now we have $v(x) = -8x^2 + 17.6x = \frac {dx}{dt}$
then plug it back into the Power function $P(t)= -80x(-8x^2 + 17.6x) + 85.3(-8x^2 + 17.6x)$, so $P(t)= 640x^3 - 1408x - 682.4x^2 + 1501.28x$, then $P(t)= 640x^3 - 682.4x^2 + 93.28x$ I plugged in $x = 50 cm = .5 m$ and did not get the right answer. So could anyone tell me if my steps even make sense and are legal? Maybe it was just a simple negative sign or mistyping of numbers in a calculator. Thank you.
The problem asks for the work done by the $100$ N pulling force. This force is always $100$ N and does not depend on distance. Thus the total work it has done after the block moves a distance $x$ is $Fx$ where $F$ is a constant $100$ N. From there, your reasoning would lead to $P = Fv$, just like the solution. If you were asked for the total work done by all the forces on the block then it would look more like your answer (though I haven't looked through it closely so I'm not sure it's exactly right in detail).