A dynamics problem uses Work–energy theorem. Given mass (m) = 2,180 kg; initial speed (v1) = 9.0000 m/s; final speed (v2) = 13.0000 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) = 2,180 kg, initial speed (v1) = 9.0000 m/s, final speed (v2) = 13.0000 m/s.

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

5. Step 5 — Evaluate:

   $$W = 95920\ \text{J}$$

6. Step 6 — Check: returning W = 95,920 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

• 191,840 — kept a factor of two that cancels in the correct rearrangement.
• 47,960 — dropped that same factor in the other direction.
• 105,512 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Potential Energy

A dynamics problem uses Work–energy theorem. Given initial speed (v1) = 14.0000 m/s; final speed (v2) = 10.5000 m/s; work done (W) = 198,345 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) = 14.0000 m/s, final speed (v2) = 10.5000 m/s, work done (W) = 198,345 J.

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

5. Step 5 — Evaluate:

   $$m = -4626\ \text{kg}$$

6. Step 6 — Check: returning m = -4,626 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

• -9,252 — kept a factor of two that cancels in the correct rearrangement.
• -2,313 — dropped that same factor in the other direction.
• -5,089 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Potential Energy

A dynamics problem uses Work–energy theorem. Given mass (m) = 300.0 kg; initial speed (v1) = 8.0000 m/s; final speed (v2) = 21.5000 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) = 300.0 kg, initial speed (v1) = 8.0000 m/s, final speed (v2) = 21.5000 m/s.

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

5. Step 5 — Evaluate:

   $$W = 59738\ \text{J}$$

6. Step 6 — Check: returning W = 59,738 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

• 119,475 — kept a factor of two that cancels in the correct rearrangement.
• 29,869 — dropped that same factor in the other direction.
• 65,711 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Potential Energy

A dynamics problem uses Work–energy theorem. Given initial speed (v1) = 13.5000 m/s; final speed (v2) = 14.5000 m/s; work done (W) = 1,581,871 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) = 13.5000 m/s, final speed (v2) = 14.5000 m/s, work done (W) = 1,581,871 J.

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

5. Step 5 — Evaluate:

   $$m = 112991\ \text{kg}$$

6. Step 6 — Check: returning m = 112,991 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

• 225,982 — kept a factor of two that cancels in the correct rearrangement.
• 56,495 — dropped that same factor in the other direction.
• 124,290 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Potential Energy

A dynamics problem uses Work–energy theorem. Given mass (m) = 480.0 kg; initial speed (v1) = 15.0000 m/s; final speed (v2) = 13.5000 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) = 480.0 kg, initial speed (v1) = 15.0000 m/s, final speed (v2) = 13.5000 m/s.

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

5. Step 5 — Evaluate:

   $$W = -10260\ \text{J}$$

6. Step 6 — Check: returning W = -10,260 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

• -20,520 — kept a factor of two that cancels in the correct rearrangement.
• -5,130 — dropped that same factor in the other direction.
• -11,286 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Potential Energy

A dynamics problem uses Work–energy theorem. Given initial speed (v1) = 6.5000 m/s; final speed (v2) = 38.5000 m/s; work done (W) = 213,346 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) = 6.5000 m/s, final speed (v2) = 38.5000 m/s, work done (W) = 213,346 J.

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

5. Step 5 — Evaluate:

   $$m = 296.3\ \text{kg}$$

6. Step 6 — Check: returning m = 296.3 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

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

Reference: FE Reference Handbook — Dynamics → Potential Energy

A dynamics problem uses Work–energy theorem. Given mass (m) = 1,050 kg; initial speed (v1) = 1.5000 m/s; final speed (v2) = 15.0000 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) = 1,050 kg, initial speed (v1) = 1.5000 m/s, final speed (v2) = 15.0000 m/s.

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

5. Step 5 — Evaluate:

   $$W = 116944\ \text{J}$$

6. Step 6 — Check: returning W = 116,944 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

• 233,888 — kept a factor of two that cancels in the correct rearrangement.
• 58,472 — dropped that same factor in the other direction.
• 128,638 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Potential Energy

A dynamics problem uses Work–energy theorem. Given initial speed (v1) = 18.5000 m/s; final speed (v2) = 14.5000 m/s; work done (W) = 462,141 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) = 18.5000 m/s, final speed (v2) = 14.5000 m/s, work done (W) = 462,141 J.

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

5. Step 5 — Evaluate:

   $$m = -7002\ \text{kg}$$

6. Step 6 — Check: returning m = -7,002 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

• -14,004 — kept a factor of two that cancels in the correct rearrangement.
• -3,501 — dropped that same factor in the other direction.
• -7,702 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Potential Energy

A dynamics problem uses Work–energy theorem. Given mass (m) = 560.0 kg; initial speed (v1) = 4.5000 m/s; final speed (v2) = 13.0000 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) = 560.0 kg, initial speed (v1) = 4.5000 m/s, final speed (v2) = 13.0000 m/s.

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

5. Step 5 — Evaluate:

   $$W = 41650\ \text{J}$$

6. Step 6 — Check: returning W = 41,650 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

• 83,300 — kept a factor of two that cancels in the correct rearrangement.
• 20,825 — dropped that same factor in the other direction.
• 45,815 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Potential Energy

A dynamics problem uses Work–energy theorem. Given initial speed (v1) = 7.5000 m/s; final speed (v2) = 37.0000 m/s; work done (W) = 1,587,798 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) = 37.0000 m/s, work done (W) = 1,587,798 J.

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

5. Step 5 — Evaluate:

   $$m = 2419\ \text{kg}$$

6. Step 6 — Check: returning m = 2,419 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

• 4,838 — kept a factor of two that cancels in the correct rearrangement.
• 1,210 — dropped that same factor in the other direction.
• 2,661 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Potential Energy