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A block (mass = 4.0 kg) sliding on a horizontal frictionless surface is attached to one end of a horizontal spring (k = 100 N/m) which has its other end fixed. If the maximum distance the block slides from the equilibrium position is equal to 20 cm, what is the speed of the block at an instant when it is a distance of 16 cm from the equilibrium position

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Answer:

The speed of the block at a distance of 16 cm from the equilibrium position is approximately 0.19 meters per second.

Step-by-step explanation:

To find the speed of the block at a specific distance from the equilibrium position, we can use the conservation of mechanical energy. The total mechanical energy of the system (block + spring) remains constant as long as there are no external forces doing work on it. The mechanical energy is the sum of the kinetic energy (KE) and the potential energy stored in the spring (PE):

Total Mechanical Energy (E) = KE + PE

At the equilibrium position (where the block is not displaced), the spring is not stretched or compressed, so the potential energy is zero, and all the mechanical energy is in the form of kinetic energy. As the block moves away from the equilibrium position, some of its kinetic energy gets converted into potential energy stored in the spring.

At a distance x from the equilibrium position, the potential energy stored in the spring is given by:

PE = 0.5 * k * x^2

And the kinetic energy of the block is given by:

KE = 0.5 * m * v^2

where:

k is the spring constant (100 N/m)

x is the displacement from the equilibrium position (16 cm = 0.16 m)

m is the mass of the block (4.0 kg)

v is the velocity of the block

We know that the total mechanical energy remains constant, so at the initial position (x = 0), all the energy is kinetic:

E_initial = KE_initial = 0.5 * m * v_initial^2

At the position with x = 0.16 m, the total mechanical energy is the sum of kinetic and potential energy:

E = KE + PE = 0.5 * m * v^2 + 0.5 * k * x^2

Since E_initial = E, we can equate the expressions for mechanical energy at these two positions:

0.5 * m * v_initial^2 = 0.5 * m * v^2 + 0.5 * k * x^2

Solving for v (velocity of the block at 16 cm from equilibrium):

v^2 = v_initial^2 + (k/m) * x^2

Substitute the known values:

v^2 = v_initial^2 + (100 N/m / 4.0 kg) * (0.16 m)^2

Now, plug in the values and calculate:

v^2 = v_initial^2 + 0.64 m^2/s^2

Given that the maximum displacement is 20 cm (0.20 m) and we want to find the velocity at 16 cm (0.16 m), we can use the concept of conservation of energy. At maximum displacement, all energy is potential energy; at 16 cm displacement, all energy is the sum of potential and kinetic energy. Thus, the potential energy difference becomes the kinetic energy at 16 cm displacement:

Potential Energy Difference = 0.5 * k * (0.20 m)^2

Kinetic Energy at 16 cm = 0.5 * k * (0.16 m)^2

Setting these two equal:

0.5 * k * (0.20 m)^2 = 0.5 * k * (0.16 m)^2 + 0.5 * m * v^2

Now we can solve for v:

0.5 * m * v^2 = 0.5 * k * (0.20 m)^2 - 0.5 * k * (0.16 m)^2

v^2 = ((0.5 * k * (0.20 m)^2 - 0.5 * k * (0.16 m)^2) / m

v = sqrt(((0.5 * k * (0.20 m)^2 - 0.5 * k * (0.16 m)^2) / m))

Now plug in the values for k, m, and calculate the velocity:

v = sqrt(((0.5 * 100 N/m * (0.20 m)^2 - 0.5 * 100 N/m * (0.16 m)^2) / 4.0 kg))

v = sqrt(((10 N * 0.04 m - 10 N * 0.0256 m) / 4.0 kg))

v = sqrt((0.4 N - 0.256 N) / 4.0 kg)

v = sqrt(0.144 N / 4.0 kg)

v = sqrt(0.036 m^2/s^2)

v = 0.19 m/s

So, the speed of the block at a distance of 16 cm from the equilibrium position is approximately 0.19 meters per second.

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