A 0.25-kg block oscillates on the end of the spring with a spring constant of 200 N/m. If the oscillation is started by elongating the spring 0.15 m and giving the block a speed of 3.0 m/s, then the maximum speed of the block is

A3.7 m/s
B0.18 m/s
C5.2 m/s
D13 m/s
E0.13 m/s

A 0.25 kg block oscillates on the end of a spring with a spring constant of 100 N/m. If the

oscillation is started by elongating the spring 0.1 m and giving the block a speed of 3 m/s,
then the amplitude of the oscillation is:

To solve this problem, we need to apply the principle of conservation of mechanical energy.

First, let's calculate the potential energy stored in the spring when it is elongated by 0.15 m. The potential energy can be calculated using the formula:

Potential Energy = 0.5 * k * x^2

where k is the spring constant and x is the displacement from the equilibrium position.

Substituting the given values, we have:

Potential Energy = 0.5 * (200 N/m) * (0.15 m)^2
Potential Energy = 0.5 * 200 N/m * 0.0225 m^2
Potential Energy = 2.25 J

Next, let's consider the initial kinetic energy of the block. The kinetic energy can be calculated using the formula:

Kinetic Energy = 0.5 * m * v^2

where m is the mass of the block and v is its velocity.

Substituting the given values, we have:

Kinetic Energy = 0.5 * (0.25 kg) * (3.0 m/s)^2
Kinetic Energy = 0.375 J

According to the principle of conservation of mechanical energy, the total mechanical energy (potential energy + kinetic energy) remains constant throughout the oscillation.

Therefore, the maximum kinetic energy of the block (and hence its maximum speed) can be calculated by subtracting the potential energy from the initial mechanical energy.

Maximum Kinetic Energy = Initial Mechanical Energy - Potential Energy
Maximum Kinetic Energy = Kinetic Energy - Potential Energy
Maximum Kinetic Energy = 0.375 J - 2.25 J
Maximum Kinetic Energy = -1.875 J

Since kinetic energy cannot be negative, we can conclude that the maximum speed of the block is 0 m/s.

Therefore, the answer is E) 0.13 m/s.

PE(max) = KE +PE

kx(max)²/2 =mv²/2 +kx²/2
x(max) = sqrt(mv²/k +x²)=
=sqrt(0.25•9/200 +0.15²)=0.184.
v(max) = ω•x(max) = x(max) •sqrt(k/m) =
=0.184•sqrt(200/0.25)=5.2 m/s

Well, this problem is a real spring-tacular! To find the maximum speed of the block, we can use the principle of conservation of mechanical energy.

When the block is at its maximum displacement (0.15 m), all of its initial potential energy (due to the elongated spring) will be converted into kinetic energy. So, we can set the potential energy equal to the kinetic energy:

(1/2)kx^2 = (1/2)mv^2

Where k is the spring constant, x is the maximum displacement, m is the mass of the block, and v is the maximum speed of the block.

Plugging in the values given, we have:

(1/2)(200 N/m)(0.15 m)^2 = (1/2)(0.25 kg)v^2

Simplifying, we get:

22.5 N = 0.03125 kgv^2

Dividing both sides by 0.03125 kg, we get:

v^2 = 720 m^2/s^2

Taking the square root of both sides, we find:

v ≈ 26.87 m/s

But hold on, that's not one of the answer choices! Looks like we got ourselves into a pickle here. Perhaps I went a little too far with my spring-loaded humor. So, let's try to figure out what went wrong.

Oh, I see it now! I made a mistake in my calculations. So, let's redo it:

(1/2)(200 N/m)(0.15 m)^2 = (1/2)(0.25 kg)v^2

Now, calculating that, we get:

4.5 N = 0.03125 kgv^2

Dividing both sides by 0.03125 kg, we get:

v^2 = 144 m^2/s^2

Taking the square root of both sides, we find:

v ≈ 12 m/s

Ha! There we go, we're back on track! So, the maximum speed of the block is approximately 12 m/s. And no, it's not D13 m/s, although that would have been fun to have a block moving faster than my joke delivery!

To find the maximum speed of the block, we need to use the principle of conservation of mechanical energy.

The total mechanical energy of the system is given by the sum of the kinetic energy (KE) and potential energy (PE):

E = KE + PE

At the maximum speed, all the potential energy is converted to kinetic energy. Therefore, the potential energy is zero, and the mechanical energy is equal to the kinetic energy.

The formula for the potential energy of a mass-spring system is:

PE = (1/2)kx^2

where k is the spring constant and x is the elongation of the spring. In this case, x = 0.15 m.

The kinetic energy of the block is given by:

KE = (1/2)mv^2

where m is the mass of the block and v is the velocity.

Setting up the equation E = KE + PE, we have:

E = KE + 0

E = (1/2)mv^2 + (1/2)kx^2

Substituting the given values:
m = 0.25 kg
v = 3.0 m/s
k = 200 N/m
x = 0.15 m

E = (1/2)(0.25 kg)(3.0 m/s)^2 + (1/2)(200 N/m)(0.15 m)^2

E = 1.125 J + 1.35 J

E = 2.475 J

Therefore, the maximum speed of the block is:

v_max = √(2E/m)

v_max = √(2(2.475 J)/(0.25 kg))

v_max = √(4.95 J/kg) = 2.23 m/s

Therefore, the correct answer is B) 0.18 m/s.