Question

A 0.20-kg mass is oscillating on a spring over a horizontal frictionless surface. When it is at a displacement of 2.6 cm for equilibrium it has a kinetic energy of 1.4 J and a spring potential energy of 2.2 J. What is the maximum speed of the mass during its oscillation?

Answer 1

Explanation:

The given data is as follows.

                    mass = 0.20 kg

              displacement = 2.6 cm

              Kinetic energy = 1.4 J

       Spring potential energy = 2.2 J

Now, we will calculate the total energy present present as follows.

         Total energy = Kinetic energy + spring potential energy

                           = 1.4 J + 2.2 J

                            = 3.6 Joules

As maximum kinetic energy of the object will be equal to the total energy.

So,      K.E = Total energy

                = 3.6 J

Also, we know that

                  K.E = $\frac{1}{2}mv^{2}_{m}$

or,                   v = $\sqrt{\frac{2K.E}{m}}$

                        = $\sqrt{2 \times 3.6 J}{0.2 kg}$

                        = $\sqrt{36}$

                        = 6 m/s

thus, we can conclude that maximum speed of the mass during its oscillation is 6 m/s.

Answer 2

The maximum speed of the mass during the oscillation is 6 m/s

Data obtained from the question:

• Mass (m) = 0.2 Kg
• Displacement (d) = 2.6 cm
• Kinetic energy (KE) = 1.4 J
• Potential energy (PE) = 2.2 J
• Maximum speed (v) =?

How to determine the maximum speed

The maximum speed during the oscillation can be obtained as follow:

• Mass (m) = 0.2 Kg
• Kinetic energy (KE) = 1.4 J
• Potential energy (PE) = 2.2 J
• Total energy (E) = 1.4 + 2.2 = 3.6 J
• Maximum speed (v) =?

E = ½mv²

3.6 = ½ × 0.2 × v²

3.6 = 0.1 × v²

Divide both side by 0.1

v² = 3.6 / 0.1

v² = 36

Take the square root of both side

v = √36

v = 6 m/s

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