A dynamics problem uses Work–energy theorem. Given mass (m) = 1,660 kg; initial speed (v1) = 5.0000 m/s; final speed (v2) = 32.0000 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) = 1,660 kg, initial speed (v1) = 5.0000 m/s, final speed (v2) = 32.0000 m/s.

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

5. Step 5 — Evaluate:

   $$W = 829170\ \text{J}$$

6. Step 6 — Check: returning W = 829,170 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

• 1,658,340 — kept a factor of two that cancels in the correct rearrangement.
• 414,585 — dropped that same factor in the other direction.
• 912,087 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Friction

A dynamics problem uses Work–energy theorem. Given initial speed (v1) = 9.5000 m/s; final speed (v2) = 24.5000 m/s; work done (W) = 1,415,621 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) = 9.5000 m/s, final speed (v2) = 24.5000 m/s, work done (W) = 1,415,621 J.

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

5. Step 5 — Evaluate:

   $$m = 5551\ \text{kg}$$

6. Step 6 — Check: returning m = 5,551 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

• 11,103 — kept a factor of two that cancels in the correct rearrangement.
• 2,776 — dropped that same factor in the other direction.
• 6,107 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Friction

A dynamics problem uses Work–energy theorem. Given mass (m) = 1,300 kg; initial speed (v1) = 2.5000 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) = 1,300 kg, initial speed (v1) = 2.5000 m/s, final speed (v2) = 20.0000 m/s.

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

5. Step 5 — Evaluate:

   $$W = 255938\ \text{J}$$

6. Step 6 — Check: returning W = 255,938 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

• 511,875 — kept a factor of two that cancels in the correct rearrangement.
• 127,969 — dropped that same factor in the other direction.
• 281,531 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Friction

A dynamics problem uses Work–energy theorem. Given initial speed (v1) = 20.0000 m/s; final speed (v2) = 17.5000 m/s; work done (W) = 466,703 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) = 20.0000 m/s, final speed (v2) = 17.5000 m/s, work done (W) = 466,703 J.

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

5. Step 5 — Evaluate:

   $$m = -9956\ \text{kg}$$

6. Step 6 — Check: returning m = -9,956 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

• -19,913 — kept a factor of two that cancels in the correct rearrangement.
• -4,978 — dropped that same factor in the other direction.
• -10,952 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Friction

A dynamics problem uses Work–energy theorem. Given mass (m) = 2,500 kg; initial speed (v1) = 5.0000 m/s; final speed (v2) = 40.0000 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) = 2,500 kg, initial speed (v1) = 5.0000 m/s, final speed (v2) = 40.0000 m/s.

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

5. Step 5 — Evaluate:

   $$W = 1968750\ \text{J}$$

6. Step 6 — Check: returning W = 1,968,750 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

• 3,937,500 — kept a factor of two that cancels in the correct rearrangement.
• 984,375 — dropped that same factor in the other direction.
• 2,165,625 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Friction

A dynamics problem uses Work–energy theorem. Given initial speed (v1) = 19.5000 m/s; final speed (v2) = 32.0000 m/s; work done (W) = 727,445 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) = 19.5000 m/s, final speed (v2) = 32.0000 m/s, work done (W) = 727,445 J.

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

5. Step 5 — Evaluate:

   $$m = 2260\ \text{kg}$$

6. Step 6 — Check: returning m = 2,260 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

• 4,520 — kept a factor of two that cancels in the correct rearrangement.
• 1,130 — dropped that same factor in the other direction.
• 2,486 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Friction

A dynamics problem uses Work–energy theorem. Given mass (m) = 2,030 kg; initial speed (v1) = 14.0000 m/s; final speed (v2) = 39.5000 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) = 2,030 kg, initial speed (v1) = 14.0000 m/s, final speed (v2) = 39.5000 m/s.

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

5. Step 5 — Evaluate:

   $$W = 1384714\ \text{J}$$

6. Step 6 — Check: returning W = 1,384,714 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

• 2,769,428 — kept a factor of two that cancels in the correct rearrangement.
• 692,357 — dropped that same factor in the other direction.
• 1,523,185 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Friction

A dynamics problem uses Work–energy theorem. Given initial speed (v1) = 9.0000 m/s; final speed (v2) = 28.0000 m/s; work done (W) = 678,298 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) = 9.0000 m/s, final speed (v2) = 28.0000 m/s, work done (W) = 678,298 J.

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

5. Step 5 — Evaluate:

   $$m = 1930\ \text{kg}$$

6. Step 6 — Check: returning m = 1,930 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

• 3,859 — kept a factor of two that cancels in the correct rearrangement.
• 964.9 — dropped that same factor in the other direction.
• 2,123 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Friction

A dynamics problem uses Work–energy theorem. Given mass (m) = 1,910 kg; initial speed (v1) = 6.5000 m/s; final speed (v2) = 14.0000 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) = 1,910 kg, initial speed (v1) = 6.5000 m/s, final speed (v2) = 14.0000 m/s.

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

5. Step 5 — Evaluate:

   $$W = 146831\ \text{J}$$

6. Step 6 — Check: returning W = 146,831 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

• 293,663 — kept a factor of two that cancels in the correct rearrangement.
• 73,416 — dropped that same factor in the other direction.
• 161,514 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Friction

A dynamics problem uses Work–energy theorem. Given initial speed (v1) = 17.5000 m/s; final speed (v2) = 38.5000 m/s; work done (W) = 1,474,353 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) = 38.5000 m/s, work done (W) = 1,474,353 J.

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

5. Step 5 — Evaluate:

   $$m = 2507\ \text{kg}$$

6. Step 6 — Check: returning m = 2,507 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

• 5,015 — kept a factor of two that cancels in the correct rearrangement.
• 1,254 — dropped that same factor in the other direction.
• 2,758 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Friction