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natta225 [31]
3 years ago
8

A mass m attached to a spring of constant k is oscillating on a frictionless surface. A second mass of mass m is dropped on top

of the first at the moment the spring is extended to the end (x = A). Which of the following is true?
A. The maximum velocity does not change.
B. The amplitude of the oscillation A decreases.
C. The frequency of oscillations f does not change.
D. The frequency of oscillations f increases.
E. The period of oscillation increases.
Physics
1 answer:
elixir [45]3 years ago
8 0

Answer:

E. The period of oscillation increases.

Explanation:

The period of oscillation is:

T = 2π√(m/k)

Frequency is the inverse of period (f = 1/T), so as period increases, frequency decreases.

Increasing the mass will increase the period and decrease the frequency.

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A car traveling at 27 m/s runs out of gas while traveling up a 8.0 ∘ slope. How far will it coast before starting to roll back d
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Answer:

Approximately 2.7 \times 10^{2}\; \rm m along the slope, assuming that no energy was lost to friction.

Explanation:

Let v denote the initial velocity of this vehicle. Let m denote the mass of this vehicle. The kinetic energy (KE) of this vehicle would have initially been:

\displaystyle \frac{1}{2}\, m \cdot v^{2}.

The gain in the gravitational potential energy (GPE) of this vehicle is proportional to the increase in its height.

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m \cdot g \cdot h.

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Since the vehicle went out of fuel, all its GPE would have been converted from KE. Assuming that no energy was converted to friction. The GPE of this vehicle would be maximal when the entirety of the KE was converted to GPE.

Hence, the maximal GPE of this vehicle would be equal to its initial KE:

\displaystyle m \cdot g \cdot h = \frac{1}{2}\, m \cdot v^{2}.

The maximum height of this vehicle would be:

\begin{aligned} h &= \frac{(1/2) \, m \cdot v^{2}}{m \cdot g}\\ &= \frac{v^{2}}{2\, g}\end{aligned}.

Given that v = 27\; \rm m\cdot s^{-1}, the maximum height of this vehicle would be:

\begin{aligned} & \frac{v^{2}}{2\, g} \\ =\; & \frac{(27\; \rm m\cdot s^{-1})^{2}}{2 \times 9.81\; \rm m\cdot s^{-2}} \\ \approx \; &37.2\; \rm m\cdot s^{-1}\end{aligned}.

Refer to the diagram attached. The distance that this vehicle traveled along the slope would be approximately:

\begin{aligned}\frac{37.2\; \rm m}{\sin(8.0^{\circ})} \approx 2.7 \times 10^{2}\; \rm m\end{aligned}.

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