A 1.1 kg block is held against a spring of force constant 2.0×10^4 N/m, compressing it a distance of 0.20 m.
How fast is the block moving after it is released and the spring pushes it away?
v=____m/s
Alright, hold on to your funny bone! The block, weighing in at a svelte 1.1 kg, is ready to show off its moves. With the spring of force constant 2.0×10^4 N/m, it's looking to gain some serious speed.
Now, assuming the block is released from rest, we're going to calculate its velocity as it gets sprung into action. First, let's use good ol' Hooke's Law to find the potential energy stored in the spring.
The potential energy stored in the spring is given by the formula:
PE = (1/2)kx^2
Where:
PE = Potential Energy
k = Force constant of the spring (2.0×10^4 N/m)
x = Compression distance (0.20 m)
Plugging in the numbers:
PE = (1/2)(2.0×10^4 N/m)(0.20 m)^2
After crunching the numbers (and, hopefully, not any funny bones), we find that the potential energy stored in the spring is:
PE ≈ 400 J
Now, since energy is conserved in this wacky universe of ours, all that potential energy is going to be transformed into kinetic energy as the block gets released. And we know:
KE = (1/2)mv^2
Where:
KE = Kinetic Energy
m = Mass of the block (1.1 kg)
v = Velocity of the block
Solving for v, we rearrange the equation:
v^2 = (2KE) / m
And substituting in the potential energy we found earlier:
v^2 = (2 * 400 J) / 1.1 kg
Hold on to your funny hats, folks! After some calculations, we find that the block's velocity after it is released is approximately:
v ≈ 19.15 m/s
So, our speedy block is zooming away at a velocity of around 19.15 m/s! Just remember, it's always a good idea to keep a safe distance from any spring-loaded acrobatics. Safety first, folks!
To find the velocity of the block after it is released and pushed away by the spring, we can use the principle of conservation of energy.
The potential energy stored in the compressed spring is converted into kinetic energy of the moving block.
The potential energy stored in the spring is given by the formula:
U = (1/2)kx^2
where U is the potential energy, k is the force constant of the spring, and x is the compression distance.
Substituting the given values into the formula:
U = (1/2) * (2.0×10^4 N/m) * (0.20 m)^2
U = 400 J
According to conservation of energy, the potential energy is equal to the kinetic energy:
U = (1/2)mv^2
where m is the mass of the block and v is its velocity.
Rearranging the equation to solve for velocity:
v^2 = (2U) / m
v^2 = (2 * 400 J) / 1.1 kg
v^2 = 800 J / 1.1 kg
v^2 = 727.27 m^2/s^2
Taking the square root of both sides to solve for v:
v = √(727.27 m^2/s^2)
v ≈ 26.98 m/s
Therefore, the velocity of the block after it is released by the spring is approximately 26.98 m/s.
To find the speed of the block after it is released and pushed away by the spring, we can use the principle of conservation of mechanical energy, which states that the total mechanical energy before the release is equal to the total mechanical energy after the release.
The total mechanical energy before the release consists of two parts: the potential energy stored in the compressed spring and the potential energy due to the gravitational force acting on the block.
The potential energy stored in the compressed spring can be calculated using the formula:
PE_spring = (1/2)kx^2
Where:
k is the force constant of the spring (2.0×10^4 N/m)
x is the distance the spring is compressed (0.20 m)
So, the potential energy stored in the compressed spring is:
PE_spring = (1/2)(2.0×10^4 N/m)(0.20 m)^2
The potential energy due to the gravitational force acting on the block is given by the formula:
PE_gravity = mgh
Where:
m is the mass of the block (1.1 kg)
g is the acceleration due to gravity (9.8 m/s^2)
h is the height from which the block was initially held
Since the block is being held against the spring, there is no height h involved. Therefore, the potential energy due to gravity is zero:
PE_gravity = 0
Thus, the total mechanical energy before the release is simply equal to the potential energy stored in the compressed spring:
Total mechanical energy before = PE_spring
After the release, the block converts all of the potential energy stored in the spring to kinetic energy. The kinetic energy of the block is given by the formula:
KE = (1/2)mv^2
Where:
m is the mass of the block (1.1 kg)
v is the speed of the block after release (what we need to find)
Since the total mechanical energy is conserved, we can equate the total mechanical energy before the release to the kinetic energy after the release:
Total mechanical energy before = KE
Substituting the formulas for potential energy and kinetic energy, we have:
(1/2)(2.0×10^4 N/m)(0.20 m)^2 = (1/2)(1.1 kg)v^2
Now we can solve for v:
(1.0×10^4 N/m)(0.04 m) = (1.1 kg)v^2
400 N/m = (1.1 kg)v^2
v^2 = (400 N/m)/(1.1 kg)
v^2 = 363.6 (m^2/s^2)
v = √(363.6 (m^2/s^2))
Calculating the square root, we find:
v ≈ 19.06 m/s
Therefore, the block will be moving at approximately 19.06 m/s after it is released and pushed away by the spring.
stored energy=1/2 k x^2
the KE gained is equal to that, solve for v.