Question

A 2.5 kg rock is released from rest at the surface of a pond 1.8 m deep. As the rock falls, a constant upward force of 5.0 N is exerted on it by water resistance. Let y=0 be at the bottom of the pond.

Required:
a. Calculate the nonconservative work, Wnc, done by water resistance on the rock, the gravitational potential energy of the system, U, the kinetic energy of the rock, K, and the total mechanical energy of the system, E, when the depth of the rock below the water's surface is 0 m.
b. Calculate the nonconservative work, Wnc, done by water resistance on the rock, the gravitational potential energy of the system, U, the kinetic energy of the rock, K, and the total mechanical energy of the system, E, when the depth of the rock below the water's surface is 0.50 m.
c. Calculate the nonconservative work, Wnc, done by water resistance on the rock, the gravitational potential energy of the system, U, the kinetic energy of the rock, K, and the total mechanical energy of the system, E, when the depth of the rock below the water's surface is 1.0 m.

Answer 1

Answer:

a) Nonconservative Work

$W_{disp} = 9\,J$

Final Gravitational Potential Energy

$U_{f} = 0\,J$

Final Translational Energy

$K_{f} = 35.131\,J$

b) Nonconservative Work

$W_{disp} = 6.5\,J$

Final Gravitational Potential Energy

$U_{f} = 12.259\,J$

Final Translational Energy

$K_{f} = 25.373\,J$

c) Nonconservative Work

$W_{disp} = 4\,J$

Final Gravitational Potential Energy

$U_{f} = 24.518\,J$

Final Translational Energy

$K_{f} = 15.614\,J$

Explanation:

The nonconservative work due to water resistance is defined by definition of work:

$W_{disp} = F\cdot (y_{o}-y_{f})$ (1)

Where:

$W_{disp}$ - Dissipate work, in joules.

$F$ - Resistance force, in newtons.

$y_{o}$ - Initial height, in meters.

$y_{f}$ - Final height, in meters.

The final gravitational potential energy ($U_{f}$), in joules, is calculated by means of the definition of gravitational potential energy:

$U_{f} = m\cdot g\cdot y_{f}$ (2)

Where:

$m$ - Mass of the rock, in kilograms.

$g$ - Gravitational acceleration, in meters per square second.

The final translational kinetic energy ($K_{f}$), in joules, is obtained by means of the Principle of Energy Conservation, Work-Energy Theorem and definitions of gravitational potential energy and translational kinetic energy:

$m\cdot g\cdot y_{o} = U_{f} + K_{f} + W_{disp}$ (3)

$K_{f} = m\cdot g\cdot y_{o} - U_{f} - W_{disp}$

Lastly, the mechanical energy of the system ($E$), in joules, is the sum of final gravitational potential energy, translational kinetic energy and dissipated work due to water resistance:

$E = U_{f} + K_{f} + W_{disp}$ (4)

Now we proceed to solve the exercise in each case:

a) Nonconservative Work ($F = 5\,N$, $y_{o} = 1.8\,m$, $y_{f} = 0\,m$)

$W_{disp} = (5\,N)\cdot (1.8\,m - 0\,m)$

$W_{disp} = 9\,J$

Final Gravitational Potential Energy ($m = 2.5\,kg$, $g = 9.807\,\frac{m}{s^{2}}$, $y_{f} = 0\,m$)

$U_{f} = (2.5\,kg) \cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (0\,m)$

$U_{f} = 0\,J$

Final Translational Energy ($m = 2.5\,kg$, $g = 9.807\,\frac{m}{s^{2}}$, $y_{o} = 1.8\,m$, $U_{f} = 0\,J$, $W_{disp} = 9\,J$)

$K_{f} = (2.5\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (1.8\,m) -0\,J-9\,J$

$K_{f} = 35.131\,J$

b) Nonconservative Work ($F = 5\,N$, $y_{o} = 1.8\,m$, $y_{f} = 0.50\,m$)

$W_{disp} = (5\,N)\cdot (1.8\,m - 0.5\,m)$

$W_{disp} = 6.5\,J$

Final Gravitational Potential Energy ($m = 2.5\,kg$, $g = 9.807\,\frac{m}{s^{2}}$, $y_{f} = 0.5\,m$)

$U_{f} = (2.5\,kg) \cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (0.5\,m)$

$U_{f} = 12.259\,J$

Final Translational Energy ($m = 2.5\,kg$, $g = 9.807\,\frac{m}{s^{2}}$, $y_{o} = 1.8\,m$, $U_{f} = 12.259\,J$, $W_{disp} = 6.5\,J$)

$K_{f} = (2.5\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (1.8\,m) -12.259\,J-6.5\,J$

$K_{f} = 25.373\,J$

c) Nonconservative Work ($F = 5\,N$, $y_{o} = 1.8\,m$, $y_{f} = 1\,m$)

$W_{disp} = (5\,N)\cdot (1.8\,m - 1\,m)$

$W_{disp} = 4\,J$

Final Gravitational Potential Energy ($m = 2.5\,kg$, $g = 9.807\,\frac{m}{s^{2}}$, $y_{f} = 1\,m$)

$U_{f} = (2.5\,kg) \cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (1\,m)$

$U_{f} = 24.518\,J$

Final Translational Energy ($m = 2.5\,kg$, $g = 9.807\,\frac{m}{s^{2}}$, $y_{o} = 1.8\,m$, $U_{f} = 24.518\,J$, $W_{disp} = 4\,J$)

$K_{f} = (2.5\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (1.8\,m) -24.518\,J-4\,J$

$K_{f} = 15.614\,J$