5.4 · Apply conservation of mechanical energy with nonconservative work

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

Work, Energy, and Power

EN PH 131 Engineering Mechanics: Dynamics — Study topic 5.4

What you will learn

• Identify the system, reference frame, initial state, and final state for an energy analysis.
• Distinguish conservative forces from nonconservative forces and determine the work done by each.
• Apply the mechanical-energy equation with nonconservative work using consistent signs and SI units.
• Check an energy result using units, direction, initial conditions, and limiting cases.

Set up the energy account

• Kinetic energy depends on speed, not the direction of velocity.
• Work is a signed scalar: a force opposing the displacement does negative work.
• Use the same two states for every energy and work term.

Separate conservative and nonconservative effects

• Friction opposing sliding usually does negative work.
• An applied force can do positive or negative work, depending on its direction relative to displacement.
• If a force is already represented through a potential-energy change, do not also count its work as nonconservative work.

Calculate work and solve for the unknown

• The unit of work and energy is the joule, equivalent to a newton-metre.
• A negative nonconservative work value reduces the system’s mechanical energy between the chosen states.
• The energy equation relates states; it does not by itself give the travel time.

Check the result

• Use the stated displacement direction to decide work signs.
• A zero-work limit should recover the conservative energy result.
• If the calculated final kinetic energy is negative, revisit the setup before reporting a speed.

Worked example

A block sliding down a rough ramp

1. Define the states and signs
   The system is the block; the ground-fixed frame is the observer’s inertial frame. State 1 is the top, where the block is at rest, and state 2 is 5.00 m down the ramp. Take displacement down the ramp as positive. The height drop is 5.00 sin 25.0°.
2. Identify energy changes and nonconservative work
   Gravity is included through the decrease in gravitational potential energy. Kinetic friction does negative work. The normal force is perpendicular to the displacement and does zero work. The normal-force magnitude is mg cos 25.0°, so the friction magnitude is μk mg cos 25.0°.
   $$T_1+V_1+W_f=T_2+V_2$$
3. Calculate the final speed
   The initial kinetic energy is zero. The potential-energy decrease is mg(5.00 sin 25.0°), and friction work is −μkmg cos 25.0°(5.00 m). Substituting these values gives a final kinetic energy of about 50.9 J and a speed of 5.04 m/s down the ramp.
   $$v_2=\sqrt{2g(5.00\sin25.0^\circ-0.180\cos25.0^\circ\cdot5.00)}=5.04\ \mathrm{m/s}$$

Worked example

A cart pulled by a position-dependent force

1. Define the system and states
   The system is the cart, observed from the ground-fixed inertial frame. State 1 is at x = 0 with zero speed; state 2 is at x = 3.00 m. Take rightward displacement as positive. Because the track is horizontal, gravitational potential energy does not change.
2. Find the work of each force
   The pull varies with position, so integrate its force component along the track. The friction force is constant and opposite the displacement. The normal force and weight are perpendicular to the horizontal displacement and do no work.
   $$W_{\mathrm{nc}}=\int_0^{3.00}(6.00+2.00x)\,dx-(1.50)(3.00)$$
3. Apply the energy equation
   The applied pull does 27.0 J of work and friction does −4.50 J, for 22.5 J net work. The cart starts from rest, so this work becomes final kinetic energy. Solving for speed gives 4.74 m/s in the positive x direction.
   $$\frac12(2.00)v_2^2=22.5\ \mathrm{J},\qquad v_2=4.74\ \mathrm{m/s}$$

Worked example

Braking a vehicle while it travels uphill

1. Set the states and positive direction
   The system is the vehicle, viewed in the ground-fixed inertial frame. State 1 is the start, with speed 20.0 m/s; state 2 is the stopping point, with zero speed. Take uphill displacement as positive and let the unknown stopping distance be s. The height gain is s sin 8.00°.
2. Account for gravity and braking
   The initial kinetic energy is spent increasing gravitational potential energy and doing work against the brakes. The braking work is −(4000 N)s. The normal force does no work because it is perpendicular to the road displacement.
   $$\frac12mv_1^2+W_{\mathrm{brake}}=mg(s\sin8.00^\circ)$$
3. Solve for stopping distance
   Insert the initial kinetic energy and braking work, then collect the terms proportional to s. The resulting distance is 42.6 m uphill. The final speed is zero by the stated stopping condition.
   $$s=\frac{\frac12(1200)(20.0)^2}{4000+(1200)(9.81)\sin8.00^\circ}=42.6\ \mathrm{m}$$

Common mistakes and how to avoid them

Lesson summary

• Define the system, ground-fixed observer, initial and final states, and positive direction before writing the energy balance.
• Represent conservative-force effects with potential energy and add nonconservative work with its algebraic sign.
• For a variable force along a path, integrate its along-path component to find work.
• Check that every energy term is in joules and that the result agrees with the stated directions and physical limits.

Check your understanding

Key terms

Mechanical energy

The sum of kinetic energy and the potential energy included in the model.

Nonconservative work

Work by forces whose effect is accounted for through work over the path, such as friction or an applied force.

Work

Energy transfer by a force acting through displacement; its sign depends on the force component along the displacement.

Inertial frame

A reference frame in which Newton’s laws apply without adding fictitious forces; here, the ground-fixed frame is treated as inertial.

Continue through EN PH 131

View the complete EN PH 131 University of Alberta EN PH 131: Engineering Mechanics: Dynamics curriculum and lessons

• 5.1 · Calculate work done by constant and variable forces
• 5.2 · Apply the particle work–energy principle
• 5.3 · Use gravitational and elastic potential energy
• 1.2 · Relate position, path length, and displacement
• 1.3 · Differentiate position to obtain velocity and acceleration
• 1.4 · Solve rectilinear motion with constant acceleration

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