A dynamics problem uses Work–energy theorem. Given mass (m) = 2,380 kg; initial speed (v1) = 9.5000 m/s; final speed (v2) = 7.5000 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) = 2,380 kg, initial speed (v1) = 9.5000 m/s, final speed (v2) = 7.5000 m/s.

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

5. Step 5 — Evaluate:

   $$W = -40460\ \text{J}$$

6. Step 6 — Check: returning W = -40,460 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

• -80,920 — kept a factor of two that cancels in the correct rearrangement.
• -20,230 — dropped that same factor in the other direction.
• -44,506 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Normal and Tangential Kinetics for Planar Problems

A dynamics problem uses Work–energy theorem. Given initial speed (v1) = 4.5000 m/s; final speed (v2) = 33.5000 m/s; work done (W) = 1,110,603 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) = 4.5000 m/s, final speed (v2) = 33.5000 m/s, work done (W) = 1,110,603 J.

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

5. Step 5 — Evaluate:

   $$m = 2016\ \text{kg}$$

6. Step 6 — Check: returning m = 2,016 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

• 4,031 — kept a factor of two that cancels in the correct rearrangement.
• 1,008 — dropped that same factor in the other direction.
• 2,217 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Normal and Tangential Kinetics for Planar Problems

A dynamics problem uses Work–energy theorem. Given mass (m) = 2,440 kg; initial speed (v1) = 7.5000 m/s; final speed (v2) = 15.5000 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) = 2,440 kg, initial speed (v1) = 7.5000 m/s, final speed (v2) = 15.5000 m/s.

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

5. Step 5 — Evaluate:

   $$W = 224480\ \text{J}$$

6. Step 6 — Check: returning W = 224,480 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

• 448,960 — kept a factor of two that cancels in the correct rearrangement.
• 112,240 — dropped that same factor in the other direction.
• 246,928 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Normal and Tangential Kinetics for Planar Problems

A dynamics problem uses Work–energy theorem. Given initial speed (v1) = 16.5000 m/s; final speed (v2) = 38.5000 m/s; work done (W) = 258,121 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) = 38.5000 m/s, work done (W) = 258,121 J.

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

5. Step 5 — Evaluate:

   $$m = 426.6\ \text{kg}$$

6. Step 6 — Check: returning m = 426.6 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

• 853.3 — kept a factor of two that cancels in the correct rearrangement.
• 213.3 — dropped that same factor in the other direction.
• 469.3 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Normal and Tangential Kinetics for Planar Problems

A dynamics problem uses Work–energy theorem. Given mass (m) = 2,380 kg; initial speed (v1) = 14.0000 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) = 2,380 kg, initial speed (v1) = 14.0000 m/s, final speed (v2) = 31.5000 m/s.

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

5. Step 5 — Evaluate:

   $$W = 947538\ \text{J}$$

6. Step 6 — Check: returning W = 947,538 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

• 1,895,075 — kept a factor of two that cancels in the correct rearrangement.
• 473,769 — dropped that same factor in the other direction.
• 1,042,291 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Normal and Tangential Kinetics for Planar Problems

A dynamics problem uses Work–energy theorem. Given initial speed (v1) = 7.5000 m/s; final speed (v2) = 19.5000 m/s; work done (W) = 413,621 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) = 19.5000 m/s, work done (W) = 413,621 J.

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

5. Step 5 — Evaluate:

   $$m = 2553\ \text{kg}$$

6. Step 6 — Check: returning m = 2,553 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

• 5,106 — kept a factor of two that cancels in the correct rearrangement.
• 1,277 — dropped that same factor in the other direction.
• 2,809 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Normal and Tangential Kinetics for Planar Problems

A dynamics problem uses Work–energy theorem. Given mass (m) = 1,290 kg; initial speed (v1) = 5.0000 m/s; final speed (v2) = 36.5000 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) = 1,290 kg, initial speed (v1) = 5.0000 m/s, final speed (v2) = 36.5000 m/s.

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

5. Step 5 — Evaluate:

   $$W = 843176\ \text{J}$$

6. Step 6 — Check: returning W = 843,176 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

• 1,686,353 — kept a factor of two that cancels in the correct rearrangement.
• 421,588 — dropped that same factor in the other direction.
• 927,494 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Normal and Tangential Kinetics for Planar Problems

A dynamics problem uses Work–energy theorem. Given initial speed (v1) = 10.0000 m/s; final speed (v2) = 30.0000 m/s; work done (W) = 50,667 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) = 10.0000 m/s, final speed (v2) = 30.0000 m/s, work done (W) = 50,667 J.

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

5. Step 5 — Evaluate:

   $$m = 126.7\ \text{kg}$$

6. Step 6 — Check: returning m = 126.7 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

• 253.3 — kept a factor of two that cancels in the correct rearrangement.
• 63.3338 — dropped that same factor in the other direction.
• 139.3 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Normal and Tangential Kinetics for Planar Problems

A dynamics problem uses Work–energy theorem. Given mass (m) = 1,300 kg; initial speed (v1) = 11.5000 m/s; final speed (v2) = 34.5000 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) = 1,300 kg, initial speed (v1) = 11.5000 m/s, final speed (v2) = 34.5000 m/s.

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

5. Step 5 — Evaluate:

   $$W = 687700\ \text{J}$$

6. Step 6 — Check: returning W = 687,700 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

• 1,375,400 — kept a factor of two that cancels in the correct rearrangement.
• 343,850 — dropped that same factor in the other direction.
• 756,470 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Normal and Tangential Kinetics for Planar Problems

A dynamics problem uses Work–energy theorem. Given initial speed (v1) = 3.0000 m/s; final speed (v2) = 22.5000 m/s; work done (W) = 423,755 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) = 22.5000 m/s, work done (W) = 423,755 J.

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

5. Step 5 — Evaluate:

   $$m = 1704\ \text{kg}$$

6. Step 6 — Check: returning m = 1,704 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

• 3,409 — kept a factor of two that cancels in the correct rearrangement.
• 852.2 — dropped that same factor in the other direction.
• 1,875 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Normal and Tangential Kinetics for Planar Problems