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kenny6666 [7]
3 years ago
14

A large fake cookie sliding on a horizontal surface is attached to one end of a horizontal spring with spring constant k = 375 N

/m; the other end of the spring is fixed in place. The cookie has a kinetic energy of 19.0 J as it passes through the position where the spring is unstretched. As the cookie slides, a frictional force of magnitude 10.0 N acts on it. (a) How far will the cookie slide from the position where the spring is unstretched before coming momentarily to rest? m (b) What will be the kinetic energy of the cookie as it slides back through the position where the spring is unstretched?
Physics
1 answer:
spayn [35]3 years ago
8 0

Answer:

The conservation of energy should be used to answer this question.

a)

At the position where the spring is unstretched, the elastic potential energy of the spring is zero.

K_1 + U_1 - W_f = K_2 +U_2\\K_1 - W_f = U_2

since U_1 and K_2 is equal to zero.

W_f = F_fx\\\\U_2 = \frac{1}{2}kx^2\\\\19 - (10)x = \frac{1}{2}(375)x^2\\\\375x^2 + 20x - 38 = 0

The roots of this quadratic equation can be solved by using discriminant.

\Delta = b^2 - 4ac\\x_{1,2} = \frac{-b \pm \sqrt{\Delta}}{2a}

x_1 = -0.346\\x_2 = 0.292

We should use the positive root, so

x = 0.292 m.

b)

We should use energy conservation between the point where the spring is momentarily at rest, and the point where the spring is unstretched.

K_2 + U_2 - W_f = K_3 + U_3\\U_2 - W_f = K_3

since the kinetic energy at point 2 and the potential energy at point 3 is equal to zero.

\frac{1}{2}kx^2 - F_fx = K_3\\K_3 = 15.987 - 2.92 = 13.067 J

Explanation:

In questions with springs, the important thing is to figure out the points where kinetic or potential energy terms would be zero. When the spring is unstretched, the elastic potential energy is zero. And when the spring is at rest, naturally the kinetic energy is equal to zero.

In part b) the cookie slides back to its original position, so the distance traveled, x, is equal to the distance in part a). The frictional force is constant in the system, so it is quite simple to solve part b) after solving part a).

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A sound source A and a reflecting surface B move directly toward each other. Relative to the air, the speed of source A is 28.7
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(a) 1440.5 Hz

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f'=(\frac{v+v_r}{v+v_s})f

where

f is the original frequency

f is the apparent frequency

v is the velocity of the wave

v_r is the velocity of the receiver (positive if the receiver is moving towards the source, negative otherwise)

v_s is the velocity of the source (positive if the source is moving away from the receiver, negative otherwise)

Here we have

f = 1110 Hz

v = 334 m/s

In the reflector frame (= on surface B), we have also

v_s = v_A = -28.7 m/s (surface A is the source, which is moving towards the receiver)

v_r = +62.2 m/s (surface B is the receiver, which is moving towards the source)

So, the frequency observed in the reflector frame is

f'=(\frac{334 m/s+62.2 m/s}{334 m/s-28.7 m/s})1110 Hz=1440.5 Hz

(b) 0.232 m

The wavelength of a wave is given by

\lambda=\frac{v}{f}

where

v is the speed of the wave

f is the frequency

In the reflector frame,

f = 1440.5 Hz

So the wavelength is

\lambda=\frac{334 m/s}{1440.5 Hz}=0.232 m

(c) 1481.2 Hz

Again, we can use the same formula

f'=(\frac{v+v_r}{v+v_s})f

In the source frame (= on surface A), we have

v_s = v_B = -62.2 m/s (surface B is now the source, since it reflects the wave, and it is moving towards the receiver)

v_r = +28.7 m/s (surface A is now the receiver, which is moving towards the source)

So, the frequency observed in the source frame is

f'=(\frac{334 m/s+28.7 m/s}{334 m/s-62.2 m/s})1110 Hz=1481.2 Hz

(d) 0.225 m

The wavelength of the wave is given by

\lambda=\frac{v}{f}

where in this case we have

v = 334 m/s

f = 1481.2 Hz is the apparent in the source frame

So the wavelength is

\lambda=\frac{334 m/s}{1481.2 Hz}=0.225 m

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