A dynamics problem uses Work–energy theorem. Given mass (m) = 650.0 kg; initial speed (v1) = 6.0000 m/s; final speed (v2) = 19.0000 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) = 650.0 kg, initial speed (v1) = 6.0000 m/s, final speed (v2) = 19.0000 m/s.

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

5. Step 5 — Evaluate:

   $$W = 105625\ \text{J}$$

6. Step 6 — Check: returning W = 105,625 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

• 211,250 — kept a factor of two that cancels in the correct rearrangement.
• 52,813 — dropped that same factor in the other direction.
• 116,188 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Potential Energy in Gravity Field

A dynamics problem uses Work–energy theorem. Given initial speed (v1) = 5.0000 m/s; final speed (v2) = 6.5000 m/s; work done (W) = 665,793 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) = 5.0000 m/s, final speed (v2) = 6.5000 m/s, work done (W) = 665,793 J.

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

5. Step 5 — Evaluate:

   $$m = 77193\ \text{kg}$$

6. Step 6 — Check: returning m = 77,193 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

• 154,387 — kept a factor of two that cancels in the correct rearrangement.
• 38,597 — dropped that same factor in the other direction.
• 84,913 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Potential Energy in Gravity Field

A dynamics problem uses Work–energy theorem. Given mass (m) = 2,060 kg; initial speed (v1) = 3.5000 m/s; final speed (v2) = 20.0000 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) = 2,060 kg, initial speed (v1) = 3.5000 m/s, final speed (v2) = 20.0000 m/s.

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

5. Step 5 — Evaluate:

   $$W = 399383\ \text{J}$$

6. Step 6 — Check: returning W = 399,383 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

• 798,765 — kept a factor of two that cancels in the correct rearrangement.
• 199,691 — dropped that same factor in the other direction.
• 439,321 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Potential Energy in Gravity Field

A dynamics problem uses Work–energy theorem. Given initial speed (v1) = 3.0000 m/s; final speed (v2) = 15.0000 m/s; work done (W) = 1,970,256 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) = 3.0000 m/s, final speed (v2) = 15.0000 m/s, work done (W) = 1,970,256 J.

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

5. Step 5 — Evaluate:

   $$m = 18243\ \text{kg}$$

6. Step 6 — Check: returning m = 18,243 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

• 36,486 — kept a factor of two that cancels in the correct rearrangement.
• 9,122 — dropped that same factor in the other direction.
• 20,067 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Potential Energy in Gravity Field

A dynamics problem uses Work–energy theorem. Given mass (m) = 1,500 kg; initial speed (v1) = 3.5000 m/s; final speed (v2) = 30.0000 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) = 1,500 kg, initial speed (v1) = 3.5000 m/s, final speed (v2) = 30.0000 m/s.

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

5. Step 5 — Evaluate:

   $$W = 665813\ \text{J}$$

6. Step 6 — Check: returning W = 665,813 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

• 1,331,625 — kept a factor of two that cancels in the correct rearrangement.
• 332,906 — dropped that same factor in the other direction.
• 732,394 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Potential Energy in Gravity Field

A dynamics problem uses Work–energy theorem. Given initial speed (v1) = 15.0000 m/s; final speed (v2) = 29.0000 m/s; work done (W) = 525,158 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) = 15.0000 m/s, final speed (v2) = 29.0000 m/s, work done (W) = 525,158 J.

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

5. Step 5 — Evaluate:

   $$m = 1705\ \text{kg}$$

6. Step 6 — Check: returning m = 1,705 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

• 3,410 — kept a factor of two that cancels in the correct rearrangement.
• 852.5 — dropped that same factor in the other direction.
• 1,876 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Potential Energy in Gravity Field

A dynamics problem uses Work–energy theorem. Given mass (m) = 1,920 kg; initial speed (v1) = 11.5000 m/s; final speed (v2) = 36.0000 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) = 1,920 kg, initial speed (v1) = 11.5000 m/s, final speed (v2) = 36.0000 m/s.

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

5. Step 5 — Evaluate:

   $$W = 1117200\ \text{J}$$

6. Step 6 — Check: returning W = 1,117,200 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

• 2,234,400 — kept a factor of two that cancels in the correct rearrangement.
• 558,600 — dropped that same factor in the other direction.
• 1,228,920 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Potential Energy in Gravity Field

A dynamics problem uses Work–energy theorem. Given initial speed (v1) = 16.5000 m/s; final speed (v2) = 24.5000 m/s; work done (W) = 781,311 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) = 16.5000 m/s, final speed (v2) = 24.5000 m/s, work done (W) = 781,311 J.

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

5. Step 5 — Evaluate:

   $$m = 4764\ \text{kg}$$

6. Step 6 — Check: returning m = 4,764 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

• 9,528 — kept a factor of two that cancels in the correct rearrangement.
• 2,382 — dropped that same factor in the other direction.
• 5,241 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Potential Energy in Gravity Field

A dynamics problem uses Work–energy theorem. Given mass (m) = 2,270 kg; initial speed (v1) = 10.0000 m/s; final speed (v2) = 31.5000 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) = 2,270 kg, initial speed (v1) = 10.0000 m/s, final speed (v2) = 31.5000 m/s.

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

5. Step 5 — Evaluate:

   $$W = 1012704\ \text{J}$$

6. Step 6 — Check: returning W = 1,012,704 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

• 2,025,408 — kept a factor of two that cancels in the correct rearrangement.
• 506,352 — dropped that same factor in the other direction.
• 1,113,974 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Potential Energy in Gravity Field

A dynamics problem uses Work–energy theorem. Given initial speed (v1) = 13.0000 m/s; final speed (v2) = 7.0000 m/s; work done (W) = 1,015,944 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) = 13.0000 m/s, final speed (v2) = 7.0000 m/s, work done (W) = 1,015,944 J.

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

5. Step 5 — Evaluate:

   $$m = -16932\ \text{kg}$$

6. Step 6 — Check: returning m = -16,932 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

• -33,865 — kept a factor of two that cancels in the correct rearrangement.
• -8,466 — dropped that same factor in the other direction.
• -18,626 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Potential Energy in Gravity Field