Consider the following experiment. You have an ideal (massless) spring with spri

Question

Consider the following experiment. You have an ideal (massless) spring with spring constant k = 64 N/m resting on a frictionless slope of angle ? = 25 degrees. Define your coordinate axis, x, as shown with x positive upwards and x = 0 as the point where the spring is not stretched or compressed.

You place an object of mass M = 5.2 kg on the slope, attach it to the spring, and slowly lower it to the point xi, where it will be at rest (stationary) with just the spring, gravitational, and normal forces acting on it. The object is connected to the spring and not moving.

You then give the object a very quick impulse so that it has a velocity vi = 2.5 m/s down the slope. The impulse is short enough that you take the initial position to be xi and the initial velocity to be vi.

If the object remains attached to the spring, what is the maximum distance, xf, that the mass goes up the slope?

Give your answer in meters to at least 3 significant digits. Do not include the units in your answer. You will not be graded on the number of significant digits, this is just so that you won't be counted incorrect due to rounding.

Hint: You will, of course, need to determine the initial position, xi, and other information

Explanation / Answer

Here we can apply the work energy concept. The final kinetic energy is zero and the initial kinetic energy = 0.5 mv^2 = 0.5*5.2*2.5^2 = 16.25 J

So change in kinetic energy = -16.25 J

Hence the net work done must be equal to -16.25 J

mgsin(theta) = k xi

hence xi = mg sin(theta)/ k = 5.2*9.8*sin(25 deg)/64 = 0.3365 m

Work done by spring = 0.5 k(xi^2-xf^2) = 0.5*64*(0.3365^2-xf^2)

Work done by gravity = -mg sin(theta)*delta_x = -5.2*9.8*sin(25 deg)*(xf+0.3365)

So we have,

0.5*64*(0.3365^2-xf^2)-5.2*9.8*sin(25 deg)*(xf+0.3365) = -16.25

3.623 - 32 xf^2 - 7.247 - 21.5366 xf = -16.25

32 xf^2 + 21.54 xf - 12.626 = 0

xf = 0.376 m = 37.6 cm

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