Question

A mass on a vertical spring begins its motion at rest at y = 0 cm. It reaches a maximum height of y = 10 cm. The two forces acting on the mass are gravity and the spring force. The graph of its kinetic energy (KE) versus position is given below. Net force on the mass varies with y as:

• A. F = 4y - 20
• B. F = 20 - 4y
• C. F = $\frac{8}{25}$(y-5)
• D. F = $\frac{8}{25}$(5-y)

Answer 1

Answer: (D)

F = $\frac{8}{25}$(5-y)

Answer 2

The problem asks us to find the net force on a mass oscillating on a vertical spring, given its kinetic energy (KE) as a function of position (y).

1. Analyze the KE-y graph:

   The graph shows that the kinetic energy (KE) is zero at y = 0 cm and y = 10 cm. This indicates that these are the turning points of the oscillation.

   The maximum kinetic energy (KE = 4 J) occurs at y = 5 cm. This is the equilibrium position where the net force is zero and the speed is maximum.

   The graph is a parabola opening downwards. Since it passes through y = 0 and y = 10, we can write the equation of the parabola in the form:

   $KE(y) = A \cdot y (y - 10)$

   To find the constant A, we use the maximum point (vertex) of the parabola, which is (y=5, KE=4).

   Substitute these values into the equation:

   $4 = A \cdot 5 (5 - 10)$

   $4 = A \cdot 5 (-5)$

   $4 = -25A$

   $A = -\frac{4}{25}$

   So, the equation for kinetic energy as a function of position y (in cm) is:

   $KE(y) = -\frac{4}{25} y (y - 10)$

   $KE(y) = -\frac{4}{25} (y^2 - 10y)$

   $KE(y) = \frac{4}{25} (10y - y^2)$

2. Relate Net Force to Kinetic Energy:

   According to the work-energy theorem, the work done by the net force is equal to the change in kinetic energy: $W_{net} = \Delta KE$.

   For a variable force, the work done is $dW_{net} = F_{net} dy$.

   Therefore, $F_{net} = \frac{d(KE)}{dy}$.

3. Calculate the Net Force:

   Differentiate the expression for KE(y) with respect to y:

   $F_{net} = \frac{d}{dy} \left[ \frac{4}{25} (10y - y^2) \right]$

   $F_{net} = \frac{4}{25} \frac{d}{dy} (10y - y^2)$

   $F_{net} = \frac{4}{25} (10 - 2y)$

   Factor out 2 from the parenthesis:

   $F_{net} = \frac{4}{25} \cdot 2 (5 - y)$

   $F_{net} = \frac{8}{25} (5 - y)$

4. Compare with Options:

   The derived expression for the net force matches option (D).

   The units of force would be J/cm if y is in cm and KE is in J. If a conversion to Newtons is expected, the factor would be $100 \times \frac{8}{25} (5-y) = 32(5-y)$ N, which is not an option. Given that option (D) is an exact match to the derived functional form, it is the intended answer.

Explanation of the solution:

1. From the given KE-position graph, identify the key points: KE=0 at y=0 and y=10 (turning points), and maximum KE=4 J at y=5 (equilibrium position).
2. The graph is a parabola. Use the roots (y=0, y=10) and the vertex (y=5, KE=4) to determine the equation of the parabola: $KE(y) = -\frac{4}{25}(y^2 - 10y) = \frac{4}{25}(10y - y^2)$.
3. The net force is the derivative of kinetic energy with respect to position: $F_{net} = \frac{d(KE)}{dy}$.
4. Differentiate the KE expression: $F_{net} = \frac{d}{dy}\left[\frac{4}{25}(10y - y^2)\right] = \frac{4}{25}(10 - 2y) = \frac{8}{25}(5 - y)$.
5. Compare this result with the given options to find the correct answer.