A block of mass 0.500 kg is pushed against a horizontal spring of negligible mass
until the spring is compressed a distance x ( see Fig. below)The force constant ofthe
spring is 450 N/rn. When it is released the block travels along a frictionless, horizontal
surface to point B, the bottom of a vertical circular track ofradius R = 1.00 m, and
continues to move along the track. The speed ofthe block at the bottom ofthe track is vB
= 12.0 m/s, and the block experiences an average friction force of 7.00 N while sliding up
the track.
a) What is x?
b) What speed do you predict for the block at the top ofthe track?
c) Does the block actually reach the top ofthe track or does it fall off before reaching the top?
a) x = 0.222 m
b) The speed of the block at the top of the track is predicted to be 11.2 m/s.
c) The block does reach the top of the track.
To answer the questions step-by-step:
a) To find x, we need to consider the potential energy stored in the spring when it is compressed. This potential energy is equal to the work done on the spring.
The formula for the potential energy stored in a spring is given by:
PE = (1/2) k x^2
where PE is the potential energy, k is the force constant of the spring, and x is the compression distance.
In this case, the potential energy is converted to kinetic energy at the bottom of the track, so we can equate the potential energy to the kinetic energy at the bottom of the track:
PE = (1/2) k x^2 = (1/2) m vB^2
Substituting the known values:
k = 450 N/m, m = 0.500 kg, and vB = 12.0 m/s:
(1/2) (450) x^2 = (1/2) (0.500) (12.0)^2
225 x^2 = 36
x^2 = 36/225
x = sqrt(36/225)
x = sqrt(0.16)
x ≈ 0.4 m
Therefore, the value of x is approximately 0.4 m.
b) To find the speed of the block at the top of the track, we need to consider the conservation of mechanical energy. At the bottom of the track, the total mechanical energy is the sum of potential energy and kinetic energy:
Ei = PE + KE
At the top of the track, the total mechanical energy is purely potential energy:
Ef = PE
Since there is no energy loss due to non-conservative forces:
Ei = Ef
(1/2) m vB^2 = m g R
Substituting the known values:
m = 0.500 kg, vB = 12.0 m/s, g = 9.8 m/s^2, and R = 1.00 m:
(1/2) (0.500) (12.0)^2 = (0.500) (9.8) (1.00)
36 = 4.9
This equation is not satisfied, which means that mechanical energy conservation is violated. As a result, the predicted speed at the top of the track cannot be determined using this approach.
c) Since the mechanical energy conservation is violated, we can conclude that the block does not reach the top of the track and it falls off before reaching it.
To find the answers to questions a, b, and c, we need to apply the laws of motion and conservation principles. Let's break down each question and find the solutions step by step.
a) What is x?
To determine the compression distance x of the spring, we can use Hooke's Law, which states that the force exerted by the spring is directly proportional to the displacement or compression of the spring.
The equation for Hooke's Law is:
F = kx
Where:
F is the force exerted by the spring
k is the force constant of the spring
x is the displacement or compression of the spring
In this case, the force exerted by the spring is equal to the weight of the block, which is given by:
F = mg
Where:
m is the mass of the block
g is the acceleration due to gravity (approximately 9.8 m/s^2)
Setting these two equations equal, we have:
mg = kx
Rearranging the equation to solve for x, we get:
x = (mg) / k
Let's plug in the values given in the problem:
m = 0.500 kg
g = 9.8 m/s^2
k = 450 N/m
x = (0.500 kg * 9.8 m/s^2) / 450 N/m
x = 0.0108 m (rounded to four decimal places)
Therefore, the compression distance of the spring is approximately 0.0108 meters.
b) What speed do you predict for the block at the top of the track?
To find the speed of the block at the top of the track, we can use the principles of conservation of mechanical energy.
At the bottom of the track, the block has both kinetic energy (1/2 mvB^2, where vB is the speed at the bottom of the track) and gravitational potential energy (mghB, where hB is the height of the bottom of the track). At the top of the track, the block will have zero kinetic energy and its entire mechanical energy will be in the form of gravitational potential energy.
Using the conservation of mechanical energy, we can equate the initial mechanical energy at the bottom of the track to the final mechanical energy at the top of the track, neglecting any losses due to friction:
(1/2 mvB^2) + mghB = mghT
Where:
vB is the speed at the bottom of the track (given as 12.0 m/s)
hB is the height at the bottom of the track (equal to the radius of the track, R = 1.00 m)
hT is the height at the top of the track (unknown)
Simplifying the equation, we get:
(1/2 vB^2) + gR = g(hT - R)
Solving for hT, we have:
hT = R + [(1/2 vB^2) / g]
Let's plug in the values given in the problem:
vB = 12.0 m/s
R = 1.00 m
g = 9.8 m/s^2
hT = 1.00 m + [(1/2 * 12.0 m/s^2 * 12.0 m/s^2) / 9.8 m/s^2]
hT ≈ 18.367 meters
Therefore, the height at the top of the track is approximately 18.367 meters.
c) Does the block actually reach the top of the track or does it fall off before reaching the top?
To determine whether the block reaches the top of the track, we need to consider the balance of forces as the block moves up the track.
The block experiences an average friction force of 7.00 N while sliding up the track. This friction force opposes the motion and causes a loss of mechanical energy. If the work done by friction is greater than or equal to the gain in gravitational potential energy, the block will fall off before reaching the top.
The work done by friction is given by:
Work_friction = friction force * distance
Where:
friction force = 7.00 N (given)
distance = hT - hB (height at the top of the track minus height at the bottom)
Substituting the values, we have:
Work_friction = (7.00 N) * (hT - R)
The gain in gravitational potential energy is given by:
Gain_energy = mghT - mghB
Substituting the values, we get:
Gain_energy = (0.500 kg) * (9.8 m/s^2) * (hT - 1.00 m)
If the work done by friction is greater than or equal to the gain in gravitational potential energy, the block will fall off before reaching the top. Otherwise, it will reach the top.
Comparing the two values, we have:
(7.00 N) * (hT - R) ≥ (0.500 kg) * (9.8 m/s^2) * (hT - 1.00 m)
By solving this inequality, we can determine whether the block reaches the top.