(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) coefficient of friction = 0.451
This was calculated by the application of energy conservation principle (the total sum of energy in a closed system is conserved)
(b) No, it comes to a stop 5.35m short of point B. This is so because the spring on expanding only does a work of 43 J on the block which is not enough to meet up the workdone of 398 J against friction.
Explanation:
The detailed step by step solution to this problems can be found in the attachment below. The solution for part (a) was divided into two: the motion of the body from point A to point B and from point B to point C. The total energy in the system is gotten from the initial gravitational potential energy. This energy becomes transformed into the work done against friction and the work done in compression the spring. A work of 398J was done in overcoming friction over a distance of 6.00m. The energy used in doing so is lost as friction is not a conservative force. This leaves only 43J of energy which compresses the spring. On expansion the spring does a work of 43J back on the block is only enough to push it over a distance of 0.65m stopping short of 5.35m from point B.
Thank you for reading and I hope this is helpful to you.
You might be interested in
Answer:
Explanation:
we know angular velocity in terms of moment of inertia and angular speed
ω .... (1)
moment of inertia of rod rotating about its center of length b
........ .(2)
using v = ωr
where w is angular velocity
and r is radius of rod which is equal to b
so we get 2v = ωb
ω = 2v/b ................. (3)
here velocity is two time because two opposite ends are moving opposite with a velocity v so net velocity will be 2v
put second and third equation in ist equation
×
so final answer will be
Data:
Centripetal Force = ? (Newton)
m (mass) = 68 Kg
s (speed) = 3.9 m/s
R (radius) = 6.5 m
Formula:
Solving:
Answer:
<span>B.159 N</span>
Centripetal Force = ? (Newton)
m (mass) = 68 Kg
s (speed) = 3.9 m/s
R (radius) = 6.5 m
Formula:
Solving:
Answer:
<span>B.159 N</span>
Answer:
Spring constant, k = 24.1 N/m
Explanation:
Given that,
Weight of the object, W = 2.45 N
Time period of oscillation of simple harmonic motion, T = 0.64 s
To find,
Spring constant of the spring.
Solution,
In case of simple harmonic motion, the time period of oscillation is given by :
m is the mass of object
m = 0.25 kg
k = 24.09 N/m
or
k = 24.11 N/m
So, the spring constant of the spring is 24.1 N/m.