(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.)
(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.)
1 answer:
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.
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Given:
Wavelength = λ = 18.7 cm
= 0.187 m
Amplitude, A = 2.34 cm
Velocity, v = 0.38 m/s
A) Calculate the angular frequency.
Angular frequency,
ω = 2π f
ω = 2π x 2.03
ω = 12.75 rad/s
B) Calculate the wave number:
C)
Since the wave is traveling in the -x direction, the sign is positive between x and t
y (x, t) = A sin(k x - ω t)
y (x, t) = 2.34 sin(33.59 x - 12.75 t)