Block A of mass 4.0 kg is on a horizontal, frictionless tabletop and is placed against a spring of negligible mass
and spring constant 650 N m. The other end of the spring is attached to a wall. The block is pushed toward the
wall until the spring has been compressed a distance x, as shown above. The block is released and follows the
trajectory shown, falling 0.80 m vertically and striking a target on the floor that is a horizontal distance of 1.2 m
from the edge of the table. Air resistance is negligible.
(a) Calculate the time elapsed from the instant block A leaves the table to the instant it strikes the floor.
(b) Calculate the speed of the block as it leaves the table.
(c) Calculate the distance x the spring was compressed.
Block B, also of mass 4.0 kg, is now placed at the edge of the table. The spring is again compressed a distance x,
and block A is released. As it nears the end of the table, it instantaneously collides with and sticks to block B.
The blocks follow the trajectory shown in the figure below and strike the floor at a horizontal distance d from the
edge of the table.
(d) Calculate d if x is equal to the value determined in part (c).
(e) Consider the system consisting of the spring, the blocks, and the table. How does the total mechanical energy
E2 of the system just before the blocks leave the table compare to the total mechanical energy E1 of the
system just before block A is released?
____ E2 < E1 ____ E2 = E1 ____ E2 > E1
Justify your answer.
E2 > E1. The total mechanical energy of the system is the sum of the potential energy stored in the spring and the kinetic energy of the blocks. When block A is released, the potential energy stored in the spring is converted to kinetic energy, so the total mechanical energy of the system increases.
(a) To calculate the time elapsed from the instant block A leaves the table to the instant it strikes the floor, we can use the equations of motion. Assuming that the only force acting on the block is gravity, we can use the equation for vertical displacement:
y = (1/2)gt^2
Where y is the vertical displacement, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time taken.
We know that the vertical displacement y is given as 0.80 m. Substituting these values into the equation, we can solve for t:
0.80 = (1/2)(9.8)t^2
0.80 = 4.9t^2
t^2 = 0.80/4.9
t^2 ≈ 0.163
Taking the square root of both sides:
t ≈ √0.163
t ≈ 0.4045 seconds
Therefore, the time elapsed from the instant block A leaves the table to the instant it strikes the floor is approximately 0.4045 seconds.
(b) To calculate the speed of the block as it leaves the table, we can use the principle of conservation of mechanical energy. The initial potential energy stored in the spring is transformed into kinetic energy as the block leaves the table, neglecting any losses due to air resistance.
The potential energy stored in the spring when it's compressed a distance x is given by:
Potential energy = (1/2)kx^2
Where k is the spring constant (650 N/m) and x is the distance the spring was compressed.
Since the potential energy is transformed into kinetic energy, we can equate the two:
Potential energy = Kinetic energy
(1/2)kx^2 = (1/2)mv^2
Where m is the mass of the block and v is its velocity.
Since the mass of the block is 4.0 kg, we can substitute the values and solve for v:
(1/2)(650)(x)^2 = (1/2)(4.0)(v)^2
Simplifying,
(650)(x)^2 = (4.0)(v)^2
[(650)(x)^2] / [(4.0)] = (v)^2
162.5(x)^2 = (v)^2
Taking the square root of both sides,
v = √[162.5(x)^2 ]
v = √[162.5(x)^2 ]
v = 12.74x
Therefore, the speed of the block as it leaves the table is 12.74 times the distance x.
(c) To calculate the distance x the spring was compressed, we can use the potential energy equation mentioned in part (b):
Potential energy = (1/2)kx^2
Where k is the spring constant (650 N/m) and x is the distance the spring was compressed.
Since the potential energy is transformed into kinetic energy, we can equate the two:
Potential energy = Kinetic energy
(1/2)kx^2 = (1/2)mv^2
Where m is the mass of the block (4.0 kg) and v is the velocity. Since the block is released from rest, the initial velocity is 0.
Therefore, we have:
(1/2)(650)(x)^2 = (1/2)(4.0)(0)^2
Simplifying,
(650)(x)^2 = 0
This implies that either x = 0 or there is an error in the problem statement. If the block was released from rest, the spring would not be compressed, and the value of x would be 0.
(d) Since x in part (c) is equal to 0, the distance d would also be 0. Therefore, the blocks would strike the floor exactly at the edge of the table.
(e) The total mechanical energy E2 of the system just before the blocks leave the table is equal to the initial potential energy stored in the spring (when it was compressed a distance x), which is given by (1/2)kx^2. The total mechanical energy E1 of the system just before block A is released is also equal to the initial potential energy stored in the spring, which is given by (1/2)kx^2. Therefore, E2 = E1. The total mechanical energy remains the same as long as there are no external forces or energy losses.
(a) To calculate the time elapsed from the instant block A leaves the table to the instant it strikes the floor, we need to consider the vertical motion of the block.
Using the equation of motion, we can find the time it takes for the block to fall a vertical distance of 0.80 m. The equation we can use is:
h = (1/2)gt^2
Where h is the vertical distance, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time.
Solving for t, we have:
t = sqrt(2h/g) = sqrt(2(0.80 m)/9.8 m/s^2) ≈ 0.40 s
So, the time elapsed from the instant the block leaves the table to the instant it strikes the floor is approximately 0.40 seconds.
(b) To calculate the speed of the block as it leaves the table, we can use the principle of conservation of mechanical energy.
The potential energy stored in the spring when it's compressed is converted into kinetic energy of the block when it is released. At the moment the block leaves the table, all of the potential energy is converted into kinetic energy.
The potential energy stored in the spring can be calculated using the formula:
PE = (1/2)kx^2
Where PE is the potential energy, k is the spring constant (650 N/m), and x is the distance the spring was compressed (which we will determine in part (c)).
The kinetic energy of the block can be calculated using the formula:
KE = (1/2)mv^2
Where KE is the kinetic energy, m is the mass of the block (4.0 kg), and v is the velocity of the block.
Since the potential energy is converted into kinetic energy, we can equate the two equations:
(1/2)kx^2 = (1/2)mv^2
Simplifying the equation, we have:
v = sqrt((k/m)x^2)
Given that k = 650 N/m and m = 4.0 kg, we can substitute these values into the equation to calculate the velocity v.
(c) To calculate the distance x the spring was compressed, we need to use the relationship between the applied force on the spring and its displacement.
According to Hooke's Law, the force applied on the spring is given by:
F = kx
Where F is the force, k is the spring constant, and x is the displacement.
Since the applied force on the spring is equal to the weight of the block (mg), we can equate the forces and solve for x:
mg = kx
Given that m = 4.0 kg and k = 650 N/m, we can substitute these values to calculate the distance x the spring was compressed in part (c).
(d) To calculate d, the horizontal distance from the edge of the table where the blocks strike the floor, we need to consider the horizontal motion of the blocks after the collision.
The horizontal distance traveled by the blocks can be calculated using the formula:
d = v*t
Where d is the distance, v is the velocity of the blocks, and t is the time taken for the blocks to fall (which we calculated in part (a)).
Given that v is the velocity of the blocks just before they leave the table, we can substitute the value we calculated in part (b) and the time from part (a) to calculate d.
(e) To compare the total mechanical energy of the system (E1) just before block A is released with the total mechanical energy of the system (E2) just before the blocks leave the table, we need to consider the conservation of mechanical energy.
In both cases, the system consists of the spring, the blocks, and the table. There are no external forces doing work on the system, and air resistance is negligible.
Therefore, according to the principle of conservation of mechanical energy, the total mechanical energy of the system remains constant.
Hence, E2 = E1.