(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.
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Answer:
B) 27.3 m
Explanation:
The rock describes a parabolic path.
The parabolic movement results from the composition of a uniform rectilinear motion (horizontal ) and a uniformly accelerated rectilinear motion of upward or downward motion (vertical ).
The equation of uniform rectilinear motion (horizontal ) for the x axis is :
x = vx*t Equation (1)
Where:
x: horizontal position in meters (m)
t : time (s)
vx: horizontal velocity in m/s
The equations of uniformly accelerated rectilinear motion of upward (vertical ) for the y axis are:
(vfy)² = (v₀y)² - 2g(y- y₀) Equation (2)
vfy = v₀y -gt Equation (3)
Where:
y: vertical position in meters (m)
y₀ : initial vertical position in meters (m)
t : time in seconds (s)
v₀y: initial vertical velocity in m/s
vfy: final vertical velocity in m/s
g: acceleration due to gravity in m/s²
Data
v₀ = 30 m/s , at an angle α=40.0° above the horizontal
v₀x = vx = 30*cos40° = 22.98 m/s
v₀y = 30*sin40° = 19.28 m/s
y₀ = 2m
y = 18.0 m
g = 9.8 m/s²
Calculation of the time (t) it takes for the rock to reach at 18 m above the ground
We replace data in the equation (2)
(vfy)² = (v₀y)² - 2g(y- y₀)
(vfy)² = (19.28)² - 2(9.8)(18- 2)
(vfy)² = 371.86 - 313.6
(vfy)² = 58.26
vfy = 7.63 m/s
We replace vfy = 7.63 m/s in the equation (2)
vfy = v₀y - gt
7.63 = 19.28 - (9.8)(t)
(9.8)(t) = 11.65
t = 11.65 / (9.8)
t = 1.19 s
Horizontal distance from where the rock was thrown to the window
We replace t = 1.19 s , in the equation (1)
x = vx*t
x = (22.98)* ( 1.19 )
x = 27.3 m
#1
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A car on the freeway experiencing a net force of -120 N.
Since net force is not zero so it will accelerate here
A submerged beach ball whose buoyant force is eight times the force of gravity on it.
Since buoyant force is more than the weight so it will not at rest and move upwards with some acceleration
#2
A frictionless spinning merry go round.
Since spinning wheel accelerating due to centripetal force so it is non inertial frame
A falling rock.
Falling rock is accelerating due to gravity so it is again non inertial frame
A hot air balloon moving at 30 degrees east of north with no net force.
since hot air balloon is having no force on it so its acceleration is zero and its an inertial frame
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Since space shuttle is accelerating due to boosters so it is non inertial frame
#3
Here net force on it is due to two given forces
towards right
towards left
so here net force is given as
since force towards left is more so the direction will be towards left
so correct answer is 60 N towards left