A dynamics problem uses Work–energy theorem. Given mass (m) = 2,070 kg; initial speed (v1) = 17.0000 m/s; final speed (v2) = 16.5000 m/s, determine the 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

1. Step 1 — State the governing relation:

   $$W = \tfrac12 m (v_2^2 - v_1^2)$$

2. Step 2 — Rearrange the relation so that W stands alone on the left-hand side.

3. Step 3 — List the givens: mass (m) = 2,070 kg, initial speed (v1) = 17.0000 m/s, final speed (v2) = 16.5000 m/s.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$W = -17336\ \text{J}$$

6. Step 6 — Check: returning W = -17,336 J to

   $$W = \tfrac12 m (v_2^2 - v_1^2)$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• -34,673 — kept a factor of two that cancels in the correct rearrangement.
• -8,668 — dropped that same factor in the other direction.
• -19,070 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Elastic Potential Energy

A dynamics problem uses Work–energy theorem. Given initial speed (v1) = 9.5000 m/s; final speed (v2) = 16.5000 m/s; work done (W) = 974,706 J, determine the 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

1. Step 1 — State the governing relation:

   $$W = \tfrac12 m (v_2^2 - v_1^2)$$

2. Step 2 — Rearrange the relation so that m stands alone on the left-hand side.

3. Step 3 — List the givens: initial speed (v1) = 9.5000 m/s, final speed (v2) = 16.5000 m/s, work done (W) = 974,706 J.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$m = 10711\ \text{kg}$$

6. Step 6 — Check: returning m = 10,711 kg to

   $$W = \tfrac12 m (v_2^2 - v_1^2)$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 21,422 — kept a factor of two that cancels in the correct rearrangement.
• 5,356 — dropped that same factor in the other direction.
• 11,782 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Elastic Potential Energy

A dynamics problem uses Work–energy theorem. Given mass (m) = 1,660 kg; initial speed (v1) = 9.5000 m/s; final speed (v2) = 11.5000 m/s, determine the 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

1. Step 1 — State the governing relation:

   $$W = \tfrac12 m (v_2^2 - v_1^2)$$

2. Step 2 — Rearrange the relation so that W stands alone on the left-hand side.

3. Step 3 — List the givens: mass (m) = 1,660 kg, initial speed (v1) = 9.5000 m/s, final speed (v2) = 11.5000 m/s.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$W = 34860\ \text{J}$$

6. Step 6 — Check: returning W = 34,860 J to

   $$W = \tfrac12 m (v_2^2 - v_1^2)$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 69,720 — kept a factor of two that cancels in the correct rearrangement.
• 17,430 — dropped that same factor in the other direction.
• 38,346 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Elastic Potential Energy

A dynamics problem uses Work–energy theorem. Given initial speed (v1) = 7.5000 m/s; final speed (v2) = 25.5000 m/s; work done (W) = 381,413 J, determine the 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

1. Step 1 — State the governing relation:

   $$W = \tfrac12 m (v_2^2 - v_1^2)$$

2. Step 2 — Rearrange the relation so that m stands alone on the left-hand side.

3. Step 3 — List the givens: initial speed (v1) = 7.5000 m/s, final speed (v2) = 25.5000 m/s, work done (W) = 381,413 J.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$m = 1284\ \text{kg}$$

6. Step 6 — Check: returning m = 1,284 kg to

   $$W = \tfrac12 m (v_2^2 - v_1^2)$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 2,568 — kept a factor of two that cancels in the correct rearrangement.
• 642.1 — dropped that same factor in the other direction.
• 1,413 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Elastic Potential Energy

A dynamics problem uses Work–energy theorem. Given mass (m) = 2,010 kg; initial speed (v1) = 18.5000 m/s; final speed (v2) = 29.5000 m/s, determine the 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

1. Step 1 — State the governing relation:

   $$W = \tfrac12 m (v_2^2 - v_1^2)$$

2. Step 2 — Rearrange the relation so that W stands alone on the left-hand side.

3. Step 3 — List the givens: mass (m) = 2,010 kg, initial speed (v1) = 18.5000 m/s, final speed (v2) = 29.5000 m/s.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$W = 530640\ \text{J}$$

6. Step 6 — Check: returning W = 530,640 J to

   $$W = \tfrac12 m (v_2^2 - v_1^2)$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 1,061,280 — kept a factor of two that cancels in the correct rearrangement.
• 265,320 — dropped that same factor in the other direction.
• 583,704 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Elastic Potential Energy

A dynamics problem uses Work–energy theorem. Given initial speed (v1) = 17.0000 m/s; final speed (v2) = 28.0000 m/s; work done (W) = 1,038,174 J, determine the 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

1. Step 1 — State the governing relation:

   $$W = \tfrac12 m (v_2^2 - v_1^2)$$

2. Step 2 — Rearrange the relation so that m stands alone on the left-hand side.

3. Step 3 — List the givens: initial speed (v1) = 17.0000 m/s, final speed (v2) = 28.0000 m/s, work done (W) = 1,038,174 J.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$m = 4195\ \text{kg}$$

6. Step 6 — Check: returning m = 4,195 kg to

   $$W = \tfrac12 m (v_2^2 - v_1^2)$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 8,389 — kept a factor of two that cancels in the correct rearrangement.
• 2,097 — dropped that same factor in the other direction.
• 4,614 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Elastic Potential Energy

A dynamics problem uses Work–energy theorem. Given mass (m) = 1,810 kg; initial speed (v1) = 14.5000 m/s; final speed (v2) = 24.5000 m/s, determine the 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

1. Step 1 — State the governing relation:

   $$W = \tfrac12 m (v_2^2 - v_1^2)$$

2. Step 2 — Rearrange the relation so that W stands alone on the left-hand side.

3. Step 3 — List the givens: mass (m) = 1,810 kg, initial speed (v1) = 14.5000 m/s, final speed (v2) = 24.5000 m/s.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$W = 352950\ \text{J}$$

6. Step 6 — Check: returning W = 352,950 J to

   $$W = \tfrac12 m (v_2^2 - v_1^2)$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 705,900 — kept a factor of two that cancels in the correct rearrangement.
• 176,475 — dropped that same factor in the other direction.
• 388,245 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Elastic Potential Energy

A dynamics problem uses Work–energy theorem. Given initial speed (v1) = 17.5000 m/s; final speed (v2) = 35.0000 m/s; work done (W) = 1,034,600 J, determine the 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

1. Step 1 — State the governing relation:

   $$W = \tfrac12 m (v_2^2 - v_1^2)$$

2. Step 2 — Rearrange the relation so that m stands alone on the left-hand side.

3. Step 3 — List the givens: initial speed (v1) = 17.5000 m/s, final speed (v2) = 35.0000 m/s, work done (W) = 1,034,600 J.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$m = 2252\ \text{kg}$$

6. Step 6 — Check: returning m = 2,252 kg to

   $$W = \tfrac12 m (v_2^2 - v_1^2)$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 4,504 — kept a factor of two that cancels in the correct rearrangement.
• 1,126 — dropped that same factor in the other direction.
• 2,477 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Elastic Potential Energy

A dynamics problem uses Work–energy theorem. Given mass (m) = 280.0 kg; initial speed (v1) = 18.0000 m/s; final speed (v2) = 22.5000 m/s, determine the 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

1. Step 1 — State the governing relation:

   $$W = \tfrac12 m (v_2^2 - v_1^2)$$

2. Step 2 — Rearrange the relation so that W stands alone on the left-hand side.

3. Step 3 — List the givens: mass (m) = 280.0 kg, initial speed (v1) = 18.0000 m/s, final speed (v2) = 22.5000 m/s.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$W = 25515\ \text{J}$$

6. Step 6 — Check: returning W = 25,515 J to

   $$W = \tfrac12 m (v_2^2 - v_1^2)$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 51,030 — kept a factor of two that cancels in the correct rearrangement.
• 12,758 — dropped that same factor in the other direction.
• 28,067 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Elastic Potential Energy

A dynamics problem uses Work–energy theorem. Given initial speed (v1) = 12.5000 m/s; final speed (v2) = 18.5000 m/s; work done (W) = 374,165 J, determine the 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

1. Step 1 — State the governing relation:

   $$W = \tfrac12 m (v_2^2 - v_1^2)$$

2. Step 2 — Rearrange the relation so that m stands alone on the left-hand side.

3. Step 3 — List the givens: initial speed (v1) = 12.5000 m/s, final speed (v2) = 18.5000 m/s, work done (W) = 374,165 J.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$m = 4023\ \text{kg}$$

6. Step 6 — Check: returning m = 4,023 kg to

   $$W = \tfrac12 m (v_2^2 - v_1^2)$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 8,047 — kept a factor of two that cancels in the correct rearrangement.
• 2,012 — dropped that same factor in the other direction.
• 4,426 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Elastic Potential Energy