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11Alexandr11
3 months ago
12

(a) A 15.0 kg block is released from rest at point A in the figure below. The track is frictionless except for the portion betwe

en points B and C, which has a length of 6.00 m. The block travels down the track, hits a spring of force constant 2,150 N/m, and compresses the spring 0.200 m from its equilibrium position before coming to rest momentarily. Determine the coefficient of kinetic friction between the block and the rough surface between points B and C.
(b) What If? The spring now expands, forcing the block back to the left. Does the block reach point B?

If the block does reach point B, how far up the curved portion of the track does it reach, and if it does not, how far short of point B does the block come to a stop? (Enter your answer in m.)

Physics
1 answer:
serg [3.5K]3 months ago
4 0

Answer:

(a) the coefficient of friction is 0.451

This was derived using the energy conservation principle (the total energy in a closed system remains constant).

(b) No, the object stops 5.35 m away from point B. This is due to the spring's expansion only performing 43 J of work on the block, which isn't sufficient compared to the 398 J required to overcome friction.

Explanation:

For more details on how this issue was resolved, refer to the attached material. The solution for part (a) separates the body’s movement into two segments: from point A to B, and from B to C. The total system energy originates from the initial gravitational potential energy, which transforms into work against friction and into work compressing the spring. A work of 398 J is needed to counteract friction over the distance of 6.00 m. The energy used for this is lost since friction is not a conservative force, leaving only 43 J for spring compression. When the spring expands, it exerts a work of 43 J back on the block, which is only sufficient to move it through a distance of 0.65 m, stopping 5.35 m short of point B.

Thank you for your attention; I trust this is beneficial to you.

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Answer:

d) v1 = v2 = v3

Explanation:

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E_i=K_i+U_i=\frac{mv_i^2}{2}+mgh_i=\frac{mv^2}{2}+mgh

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E_f=K_f+U_f=\frac{mv_f^2}{2}+mgh_f=\frac{mv_f^2}{2}

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4 0
3 months ago
When a mass of 25 g is attached to a certain spring, it makes 20 complete vibrations in 4.0 s. what is the spring constant of th
kicyunya [3294]

Response: The spring constant is 25 N/m.

Details:

The body’s mass is 25 g, which converts to 0.025 kg (since 1 kg = 1000 g).

The total oscillations are 20 in 4 seconds.

Oscillations per second = \frac{20}{4}=5

Spring's frequency of vibration is = 5 s^{-1}=5 Hz

The spring constant 'k' can be derived from the relationship involving frequency, mass, and spring constant.

Frequency=\frac{1}{2\pi}\times \sqrt{\frac{k}{m}}

5 s^{-1}=\frac{1}{2\times 3.14}\times \sqrt{\frac{k}{0.025 kg}}

k=24.649 N/m\approx 25 N/m

The spring constant is 25 N/m.

3 0
3 months ago
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Answer:3.87*10^-4

Explanation:

To determine the mass reduction, delta mass Xe, of the xenon nucleus due to its decay, we first use the provided wavelength of the gamma radiation to calculate its frequency via c = freq*wavelength.

From C=f*lambda we set up: 3*10^8=f*3.44*10^-12.

Solving gives frequency F=0.87*10^20 Hz.

Next, we calculate the emitted energy using the equation E=hf, which translates to E=f*Planck's constant.

Thus, E=0.87*10^20*6.62*10^-34, resulting in E=575.94*10^(-16).

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Utilizing the formula E=mc^2, with c^2 = 931.5 MeV/u, enables us to find the reduction in mass, yielding

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Sally is pushing a shopping cart with a force of 20 N. Because the wheels are stuck, the friction caused by the ground is exerti
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The resultant force acting on the shopping cart is 12 N to the right.

This is a scenario related to Newton's Laws. A Free Body Diagram can be utilized to illustrate all the forces acting on the object. There are four forces at play:

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The first two forces (1 and 2) counteract one another as they are equal, while the overall force can be determined by combining forces (3) and (4)

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Keith_Richards [3271]

Answer:

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