Principle of Work and Energy

Dynamics · FE Reference Handbook section

Dynamics

3 formulas

10 exam-style examples

~51 min

All Dynamics lectures

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

If Ti and Vi are, respectively, the kinetic and potential energy of a particle at state i, then for conservative systems (no energy
dissipation or gain), the law of conservation of energy is
If nonconservative forces are present, then the work done by these forces must be accounted for. Hence
computations to correctly compute the algebraic sign of the work term. If the forces serve to increase the energy of the system,

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Work–energy theorem — solve for work done — Principle of Work and Energy

A dynamics problem uses Work–energy theorem. Given mass (m) = 1,560 kg; initial speed (v1) = 19.0000 m/s; final speed (v2) = 24.5000 m/s, determine the work done (W) in J.

Given

$mass (m) = 1,560 kg$
$initial speed (v_{1}) = 19.0000 m/s$
$final speed (v_{2}) = 24.5000 m/s$

Find

work done (W), in J

Start with the thinking

The governing relation printed in this handbook section is Work–energy theorem.
Everything except W is given, so isolate W symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Dynamics items reward recognising the unknown before touching a calculator.

Step-by-step solution

Step 1 — State the governing relation:
$W = \tfrac12 m (v_2^2 - v_1^2)$
Step 2 — Rearrange the relation so that W stands alone on the left-hand side.
Step 3 — List the givens: mass (m) = 1,560 kg, initial speed (v1) = 19.0000 m/s, final speed (v2) = 24.5000 m/s.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$W = 186615\ \text{J}$
Step 6 — Check: returning W = 186,615 J to
$W = \tfrac12 m (v_2^2 - v_1^2)$
reproduces the given quantities, and both sides carry the same units.

Answer:

W = 186615\ \text{J}

Why the other options are there

373,230 — kept a factor of two that cancels in the correct rearrangement.
93,308 — dropped that same factor in the other direction.
205,277 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Principle of Work and Energy

Example 2

Work–energy theorem — solve for mass — Principle of Work and Energy (2)

A dynamics problem uses Work–energy theorem. Given initial speed (v1) = 7.0000 m/s; final speed (v2) = 28.0000 m/s; work done (W) = 1,957,113 J, determine the mass (m) in kg.

Given

$initial speed (v_{1}) = 7.0000 m/s$
$final speed (v_{2}) = 28.0000 m/s$
$work done (W) = 1,957,113 J$

Find

mass (m), in kg

Start with the thinking

The governing relation printed in this handbook section is Work–energy theorem.
Everything except m is given, so isolate m symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Dynamics items reward recognising the unknown before touching a calculator.

Step-by-step solution

Step 1 — State the governing relation:
$W = \tfrac12 m (v_2^2 - v_1^2)$
Step 2 — Rearrange the relation so that m stands alone on the left-hand side.
Step 3 — List the givens: initial speed (v1) = 7.0000 m/s, final speed (v2) = 28.0000 m/s, work done (W) = 1,957,113 J.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$m = 5325\ \text{kg}$
Step 6 — Check: returning m = 5,325 kg to
$W = \tfrac12 m (v_2^2 - v_1^2)$
reproduces the given quantities, and both sides carry the same units.

Answer:

m = 5325\ \text{kg}

Why the other options are there

10,651 — kept a factor of two that cancels in the correct rearrangement.
2,663 — dropped that same factor in the other direction.
5,858 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Principle of Work and Energy

Example 3

Work–energy theorem — solve for work done (case 2) — Principle of Work and Energy (3)

A dynamics problem uses Work–energy theorem. Given mass (m) = 950.0 kg; initial speed (v1) = 16.5000 m/s; final speed (v2) = 18.5000 m/s, determine the work done (W) in J.

Given

$mass (m) = 950.0 kg$
$initial speed (v_{1}) = 16.5000 m/s$
$final speed (v_{2}) = 18.5000 m/s$

Find

work done (W), in J

Start with the thinking

The governing relation printed in this handbook section is Work–energy theorem.
Everything except W is given, so isolate W symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Dynamics items reward recognising the unknown before touching a calculator.

Step-by-step solution

Step 1 — State the governing relation:
$W = \tfrac12 m (v_2^2 - v_1^2)$
Step 2 — Rearrange the relation so that W stands alone on the left-hand side.
Step 3 — List the givens: mass (m) = 950.0 kg, initial speed (v1) = 16.5000 m/s, final speed (v2) = 18.5000 m/s.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$W = 33250\ \text{J}$
Step 6 — Check: returning W = 33,250 J to
$W = \tfrac12 m (v_2^2 - v_1^2)$
reproduces the given quantities, and both sides carry the same units.

Answer:

W = 33250\ \text{J}

Why the other options are there

66,500 — kept a factor of two that cancels in the correct rearrangement.
16,625 — dropped that same factor in the other direction.
36,575 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Principle of Work and Energy

Example 4

Work–energy theorem — solve for mass (case 2) — Principle of Work and Energy (4)

A dynamics problem uses Work–energy theorem. Given initial speed (v1) = 8.0000 m/s; final speed (v2) = 35.5000 m/s; work done (W) = 537,711 J, determine the mass (m) in kg.

Given

$initial speed (v_{1}) = 8.0000 m/s$
$final speed (v_{2}) = 35.5000 m/s$
$work done (W) = 537,711 J$

Find

mass (m), in kg

Start with the thinking

The governing relation printed in this handbook section is Work–energy theorem.
Everything except m is given, so isolate m symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Dynamics items reward recognising the unknown before touching a calculator.

Step-by-step solution

Step 1 — State the governing relation:
$W = \tfrac12 m (v_2^2 - v_1^2)$
Step 2 — Rearrange the relation so that m stands alone on the left-hand side.
Step 3 — List the givens: initial speed (v1) = 8.0000 m/s, final speed (v2) = 35.5000 m/s, work done (W) = 537,711 J.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$m = 899.0\ \text{kg}$
Step 6 — Check: returning m = 899.0 kg to
$W = \tfrac12 m (v_2^2 - v_1^2)$
reproduces the given quantities, and both sides carry the same units.

Answer:

m = 899.0\ \text{kg}

Why the other options are there

1,798 — kept a factor of two that cancels in the correct rearrangement.
449.5 — dropped that same factor in the other direction.
988.9 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Principle of Work and Energy

Example 5

Work–energy theorem — solve for work done (case 3) — Principle of Work and Energy (5)

A dynamics problem uses Work–energy theorem. Given mass (m) = 2,590 kg; initial speed (v1) = 5.5000 m/s; final speed (v2) = 18.5000 m/s, determine the work done (W) in J.

Given

$mass (m) = 2,590 kg$
$initial speed (v_{1}) = 5.5000 m/s$
$final speed (v_{2}) = 18.5000 m/s$

Find

work done (W), in J

Start with the thinking

The governing relation printed in this handbook section is Work–energy theorem.
Everything except W is given, so isolate W symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Dynamics items reward recognising the unknown before touching a calculator.

Step-by-step solution

Step 1 — State the governing relation:
$W = \tfrac12 m (v_2^2 - v_1^2)$
Step 2 — Rearrange the relation so that W stands alone on the left-hand side.
Step 3 — List the givens: mass (m) = 2,590 kg, initial speed (v1) = 5.5000 m/s, final speed (v2) = 18.5000 m/s.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$W = 404040\ \text{J}$
Step 6 — Check: returning W = 404,040 J to
$W = \tfrac12 m (v_2^2 - v_1^2)$
reproduces the given quantities, and both sides carry the same units.

Answer:

W = 404040\ \text{J}

Why the other options are there

808,080 — kept a factor of two that cancels in the correct rearrangement.
202,020 — dropped that same factor in the other direction.
444,444 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Principle of Work and Energy

Example 6

Work–energy theorem — solve for mass (case 3) — Principle of Work and Energy (6)

A dynamics problem uses Work–energy theorem. Given initial speed (v1) = 8.5000 m/s; final speed (v2) = 13.0000 m/s; work done (W) = 231,840 J, determine the mass (m) in kg.

Given

$initial speed (v_{1}) = 8.5000 m/s$
$final speed (v_{2}) = 13.0000 m/s$
$work done (W) = 231,840 J$

Find

mass (m), in kg

Start with the thinking

The governing relation printed in this handbook section is Work–energy theorem.
Everything except m is given, so isolate m symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Dynamics items reward recognising the unknown before touching a calculator.

Step-by-step solution

Step 1 — State the governing relation:
$W = \tfrac12 m (v_2^2 - v_1^2)$
Step 2 — Rearrange the relation so that m stands alone on the left-hand side.
Step 3 — List the givens: initial speed (v1) = 8.5000 m/s, final speed (v2) = 13.0000 m/s, work done (W) = 231,840 J.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$m = 4793\ \text{kg}$
Step 6 — Check: returning m = 4,793 kg to
$W = \tfrac12 m (v_2^2 - v_1^2)$
reproduces the given quantities, and both sides carry the same units.

Answer:

m = 4793\ \text{kg}

Why the other options are there

9,585 — kept a factor of two that cancels in the correct rearrangement.
2,396 — dropped that same factor in the other direction.
5,272 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Principle of Work and Energy

Example 7

Work–energy theorem — solve for work done (case 4) — Principle of Work and Energy (7)

A dynamics problem uses Work–energy theorem. Given mass (m) = 2,520 kg; initial speed (v1) = 13.5000 m/s; final speed (v2) = 34.0000 m/s, determine the work done (W) in J.

Given

$mass (m) = 2,520 kg$
$initial speed (v_{1}) = 13.5000 m/s$
$final speed (v_{2}) = 34.0000 m/s$

Find

work done (W), in J

Start with the thinking

The governing relation printed in this handbook section is Work–energy theorem.
Everything except W is given, so isolate W symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Dynamics items reward recognising the unknown before touching a calculator.

Step-by-step solution

Step 1 — State the governing relation:
$W = \tfrac12 m (v_2^2 - v_1^2)$
Step 2 — Rearrange the relation so that W stands alone on the left-hand side.
Step 3 — List the givens: mass (m) = 2,520 kg, initial speed (v1) = 13.5000 m/s, final speed (v2) = 34.0000 m/s.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$W = 1226925\ \text{J}$
Step 6 — Check: returning W = 1,226,925 J to
$W = \tfrac12 m (v_2^2 - v_1^2)$
reproduces the given quantities, and both sides carry the same units.

Answer:

W = 1226925\ \text{J}

Why the other options are there

2,453,850 — kept a factor of two that cancels in the correct rearrangement.
613,463 — dropped that same factor in the other direction.
1,349,618 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Principle of Work and Energy

Example 8

Work–energy theorem — solve for mass (case 4) — Principle of Work and Energy (8)

A dynamics problem uses Work–energy theorem. Given initial speed (v1) = 5.0000 m/s; final speed (v2) = 18.0000 m/s; work done (W) = 1,608,040 J, determine the mass (m) in kg.

Given

$initial speed (v_{1}) = 5.0000 m/s$
$final speed (v_{2}) = 18.0000 m/s$
$work done (W) = 1,608,040 J$

Find

mass (m), in kg

Start with the thinking

The governing relation printed in this handbook section is Work–energy theorem.
Everything except m is given, so isolate m symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Dynamics items reward recognising the unknown before touching a calculator.

Step-by-step solution

Step 1 — State the governing relation:
$W = \tfrac12 m (v_2^2 - v_1^2)$
Step 2 — Rearrange the relation so that m stands alone on the left-hand side.
Step 3 — List the givens: initial speed (v1) = 5.0000 m/s, final speed (v2) = 18.0000 m/s, work done (W) = 1,608,040 J.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$m = 10756\ \text{kg}$
Step 6 — Check: returning m = 10,756 kg to
$W = \tfrac12 m (v_2^2 - v_1^2)$
reproduces the given quantities, and both sides carry the same units.

Answer:

m = 10756\ \text{kg}

Why the other options are there

21,512 — kept a factor of two that cancels in the correct rearrangement.
5,378 — dropped that same factor in the other direction.
11,832 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Principle of Work and Energy

Example 9

Work–energy theorem — solve for work done (case 5) — Principle of Work and Energy (9)

A dynamics problem uses Work–energy theorem. Given mass (m) = 1,960 kg; initial speed (v1) = 15.5000 m/s; final speed (v2) = 13.5000 m/s, determine the work done (W) in J.

Given

$mass (m) = 1,960 kg$
$initial speed (v_{1}) = 15.5000 m/s$
$final speed (v_{2}) = 13.5000 m/s$

Find

work done (W), in J

Start with the thinking

The governing relation printed in this handbook section is Work–energy theorem.
Everything except W is given, so isolate W symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Dynamics items reward recognising the unknown before touching a calculator.

Step-by-step solution

Step 1 — State the governing relation:
$W = \tfrac12 m (v_2^2 - v_1^2)$
Step 2 — Rearrange the relation so that W stands alone on the left-hand side.
Step 3 — List the givens: mass (m) = 1,960 kg, initial speed (v1) = 15.5000 m/s, final speed (v2) = 13.5000 m/s.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$W = -56840\ \text{J}$
Step 6 — Check: returning W = -56,840 J to
$W = \tfrac12 m (v_2^2 - v_1^2)$
reproduces the given quantities, and both sides carry the same units.

Answer:

W = -56840\ \text{J}

Why the other options are there

-113,680 — kept a factor of two that cancels in the correct rearrangement.
-28,420 — dropped that same factor in the other direction.
-62,524 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Principle of Work and Energy

Example 10

Work–energy theorem — solve for mass (case 5) — Principle of Work and Energy (10)

A dynamics problem uses Work–energy theorem. Given initial speed (v1) = 1.5000 m/s; final speed (v2) = 20.0000 m/s; work done (W) = 668,742 J, determine the mass (m) in kg.

Given

$initial speed (v_{1}) = 1.5000 m/s$
$final speed (v_{2}) = 20.0000 m/s$
$work done (W) = 668,742 J$

Find

mass (m), in kg

Start with the thinking

The governing relation printed in this handbook section is Work–energy theorem.
Everything except m is given, so isolate m symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Dynamics items reward recognising the unknown before touching a calculator.

Step-by-step solution

Step 1 — State the governing relation:
$W = \tfrac12 m (v_2^2 - v_1^2)$
Step 2 — Rearrange the relation so that m stands alone on the left-hand side.
Step 3 — List the givens: initial speed (v1) = 1.5000 m/s, final speed (v2) = 20.0000 m/s, work done (W) = 668,742 J.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$m = 3363\ \text{kg}$
Step 6 — Check: returning m = 3,363 kg to
$W = \tfrac12 m (v_2^2 - v_1^2)$
reproduces the given quantities, and both sides carry the same units.

Answer:

m = 3363\ \text{kg}

Why the other options are there

6,725 — kept a factor of two that cancels in the correct rearrangement.
1,681 — dropped that same factor in the other direction.
3,699 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Principle of Work and Energy