The kinetic energy of a moving object is given by the formula K=(1/2)mv^2 where m is the mass of the object and v is the velocity.

Question

The kinetic energy of a moving object is given by the formula K=(1/2)mv^2 where m is the mass of the object and v is the velocity. A 62 kg man jumps off a diving board. When he is falling at a rate of v= -0.5 m/sec, his acceleration will be -9.8 m/sec^2. How is his kinetic energy changing at that moment?

Answer 1

Using conservation of energy:

KE+PE = constant.

So, the rate of change of the kinetic energy should be equal and opposite to the rate of change of gravitational potential energy.

Gravitational potential energy = mg*y. Mass doesn't change, g doesn't change, y changes at a rate of -0.5m/s at that instant.

Therefore, PE is changing at a rate of 62kg*9.8m/s²*-0.5m/s = -303.8 kgm²/s³ = -303.8 J/s. That is to say, PE is decreasing at a rate of 303.8 J/s.

Thus, KE is increasing at a rate of 303.8 J/s.

Using calculus:

KE = mv²/2

d/dt KE = 2*m*v*(dv/dt)/2 = 2 * 62kg * -0.5m/s * -9.8m/s / 2 = 303.8 kgm²/s³ = 303.8 J/s

So the KE is increasing at a rate of 303.8 J/s.

Answer 2

1. Kinetic Energy Formula

The kinetic energy (K) of an object is given by the formula:

K = (1/2) m v²

where:

- K is the kinetic energy (in Joules, J),

- m is the mass of the object (in kilograms, kg),

- v is the velocity of the object (in meters per second, m/s).

2. Given Information

In this problem:

- Mass, m = 62 kg

- Velocity, v = -0.5 m/s (the negative sign indicates direction, but kinetic energy, being scalar, remains positive).

- Acceleration, a = -9.8 m/s²

3. Calculating Instantaneous Kinetic Energy

The kinetic energy at that instant can be calculated as:

K = (1/2) * m * v²

Substituting the values:

K = (1/2) * 62 kg * (-0.5 m/s)²

K = (1/2) * 62 kg * 0.25 m²/s²

K = (1/2) * 15.5 kg·m²/s²

K = 7.75 J

Thus, the kinetic energy at that moment is 7.75 Joules.

4. Rate of Change of Kinetic Energy

To determine how the kinetic energy is changing over time, we need to find the time derivative of kinetic energy (dK/dt). Using calculus, the derivative of kinetic energy with respect to time is given by:

dK/dt = m * v * (dv/dt)

Here:

- v = -0.5 m/s (velocity)

- dv/dt = a = -9.8 m/s² (acceleration)

5. Calculation

Substituting the values:

dK/dt = 62 kg * (-0.5 m/s) * (-9.8 m/s²)

dK/dt = 62 kg * 4.9 m²/s³

dK/dt = 303.8 W

The rate of change of kinetic energy is 303.8 Watts (W), indicating that the man's kinetic energy is increasing at this rate at that moment.

6. Interpretation

The positive rate of change of kinetic energy (303.8 W) indicates that the kinetic energy is increasing. This aligns with the fact that the man is accelerating downward under the influence of gravity, increasing his velocity and, consequently, his kinetic energy.