Physics, 09.01.2020 05:31 Ashleyvasquez2261

A4.0kg block is suspended from a spring with force constant
of500n/m. a 50g bullet
is fired into the block from directly below with a speed of
150m/sand is imbedded in the
block. (a) find the amplitude of the resulting simple
harmonicmotion. (b) what fraction
of the original kinetic energy of the bullet appears as
mechanicalenergy in the harmonic
oscillation?

Answers: 2

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Answers

Answer from: yugbug44owwc7w

(a) Amplitude of the motion is 0.184 m.

(b) 0.285 of the original kinetic energy appears as the mechanical energy in the harmonic oscillation.

Explanation:

(a) First we will use the conservation of linear momentum to find the velocity of the combined objects.

$P_1 = P_2\\mv_1 = (M + m) v_2\\0.05*150 = (4 + 0.05)v_2\\v_2 = 1.85~m/s$

Before the collision, the spring is in equilibrium, so it is stretched by the amount of x.

$kx = Mg\\x = \frac{Mg}{k} = \frac{4*9.8}{500} = 0.0784~m$

Just after the collision, the motion starts from this position.

The simple harmonic motion equation is

$x(t) = A\cos(\omega t + \phi)\\v(t) = \frac{dx(t)}{dt} = -\omega A\sin(\omega t + \phi)$

ω is the angular frequency, and given as

$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{500}{4.05}} = 11.11$

Φ is the phase angle and can be found by the initial conditions.

At t = 0 (Just after the collision);

$x(t=0) = 0.0784 = A\cos(\omega * (0) + \phi) = A\cos(\phi)\\v(t=0) = 1.85 = -(11.11)A\sin(\phi)\\\\\cos(\phi) = 0.0784/A\\A = -0.1665/\sin(\phi)\\\\\cos(\phi) = \frac{0.0784}{-\frac{0.1665}{\sin(\phi)}} = 0.4709\sin(\phi)\\\cot(\phi) = 0.4709\\\phi = 64.7^\circ\\A = \frac{0.1665}{\sin(\phi)} = \frac{0.1665}{0.9044} = 0.184~m$

(b) The original kinetic energy is

$K_0 = \frac{1}{2}mv^2 = 562.5~J$

The mechanical energy of the simple harmonic motion is

$K + U = \frac{1}{2}(M+m)v_2^2 + \frac{1}{2}k_x^2 = \frac{1}{2}(4.05)(1.85)^2 + \frac{1}{2}500(0.0784)^2\\K+ U = 160.59~J$

So 0.285 of the original kinetic energy appears as the mechanical energy in the harmonic oscillation. The rest of the original energy is converted to heat during the collision.

Answer from: Quest

b) returns freshwater to the earth.

c) returns freshwater to the earth.c) occurs after condensation.

d) returns freshwater to the earth.c) occurs after condensation.d) roughly equal to the amount of evaporation.

happy to

pls mark as brainliest.

Answer from: Quest

air exerts forces on falling objects near earth’s surface.

explanation:

the term 'free falls' refer to object that are falling towards the earth, in a situation in which gravity is the only force acting on the object.

for an object in free fall, the acceleration is constant (g, the gravitational acceleration), and so the velocity of the object constantly increases.

for objects near earth's surface, there is always another force acting on the object: the resistance due to the air. this force is opposed to the direction of fall of the object, and it is proportional to the velocity of the object, so as the object falls down, this force increases up to a point when it balances the force of gravity, and the object continues its motion at constant velocity (called terminal velocity). since in this case gravity is not the only force acting on the object, we are not in a situation of free fall.

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A4.0kg block is suspended from a spring with force constant
of500n/m. a 50g bullet
is fi...

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