( a ) ( b ) In Figure (a), a block of mass m lies on a horizontal frictionless s

Question

(a)
(b) In Figure (a), a block of mass m lies on a horizontal frictionless surface and is attached to one end of a horizontal spring (spring constant k) whose other end is fixed. The block is initially at rest at the position where the spring is unstretched (x = 0) when a constant horizontal force F rightarrow in the positive direction of the x axis is applied to it. A plot of the resulting kinetic energy of the block versus its position x is shown in Figure (b). The scale of the figure's vertical axis is set by Ks = 4.0 J and the scale of the figure's horizontal axis is set by xmax = 4.0 m.

Explanation / Answer

We are told that the net work done by the constant force F and the the spring is equal to the change in kinetic energy, K. The net work done by these forces is equal to

Wnet = Fx - (1/2)*k*x^2 = KE

That this is an inverted parabola, much like the graph of KE vs. x given in the figure. From this, we can obtain the values of F and k.

First note that the graph of KE intersects 0 J at x = 0 and x = 4

So from the equation above, we can factor out x to get:

x[F - (1/2)*k*x] = KE

Set this equal to zero and we get two solutions like expected:

x = 0 and x = 2F/k. The last one must equal 6 so that F = 2*k.

k = 4/2 =2

Then we also know that the equation for KE vs. x reaches a maximum of KE = 4.0 J at x = 2Plugging these values and the result that F = 2*k in the KE equation yields:

2[2k - (1/2)*k*2] = 4.0 --> k*(1)*(2) = 4.0---->k=2

so F = 2*2 = 4N.

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

---

---

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