A dynamics problem uses Work–energy theorem. Given mass (m) = 2,830 kg; initial speed (v1) = 10.0000 m/s; final speed (v2) = 26.5000 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) = 2,830 kg, initial speed (v1) = 10.0000 m/s, final speed (v2) = 26.5000 m/s.

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

5. Step 5 — Evaluate:

   $$W = 852184\ \text{J}$$

6. Step 6 — Check: returning W = 852,184 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

• 1,704,368 — kept a factor of two that cancels in the correct rearrangement.
• 426,092 — dropped that same factor in the other direction.
• 937,402 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Kinetics of a Rigid Body

A dynamics problem uses Work–energy theorem. Given initial speed (v1) = 16.0000 m/s; final speed (v2) = 12.0000 m/s; work done (W) = 1,851,485 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) = 16.0000 m/s, final speed (v2) = 12.0000 m/s, work done (W) = 1,851,485 J.

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

5. Step 5 — Evaluate:

   $$m = -33062\ \text{kg}$$

6. Step 6 — Check: returning m = -33,062 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

• -66,124 — kept a factor of two that cancels in the correct rearrangement.
• -16,531 — dropped that same factor in the other direction.
• -36,368 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Kinetics of a Rigid Body

A dynamics problem uses Work–energy theorem. Given mass (m) = 540.0 kg; initial speed (v1) = 8.5000 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) = 540.0 kg, initial speed (v1) = 8.5000 m/s, final speed (v2) = 7.5000 m/s.

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

5. Step 5 — Evaluate:

   $$W = -4320\ \text{J}$$

6. Step 6 — Check: returning W = -4,320 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

• -8,640 — kept a factor of two that cancels in the correct rearrangement.
• -2,160 — dropped that same factor in the other direction.
• -4,752 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Kinetics of a Rigid Body

A dynamics problem uses Work–energy theorem. Given initial speed (v1) = 15.5000 m/s; final speed (v2) = 8.0000 m/s; work done (W) = 1,227,514 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) = 15.5000 m/s, final speed (v2) = 8.0000 m/s, work done (W) = 1,227,514 J.

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

5. Step 5 — Evaluate:

   $$m = -13929\ \text{kg}$$

6. Step 6 — Check: returning m = -13,929 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

• -27,858 — kept a factor of two that cancels in the correct rearrangement.
• -6,965 — dropped that same factor in the other direction.
• -15,322 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Kinetics of a Rigid Body

A dynamics problem uses Work–energy theorem. Given mass (m) = 920.0 kg; initial speed (v1) = 9.5000 m/s; final speed (v2) = 32.5000 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) = 920.0 kg, initial speed (v1) = 9.5000 m/s, final speed (v2) = 32.5000 m/s.

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

5. Step 5 — Evaluate:

   $$W = 444360\ \text{J}$$

6. Step 6 — Check: returning W = 444,360 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

• 888,720 — kept a factor of two that cancels in the correct rearrangement.
• 222,180 — dropped that same factor in the other direction.
• 488,796 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Kinetics of a Rigid Body

A dynamics problem uses Work–energy theorem. Given initial speed (v1) = 6.0000 m/s; final speed (v2) = 7.5000 m/s; work done (W) = 1,449,522 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) = 6.0000 m/s, final speed (v2) = 7.5000 m/s, work done (W) = 1,449,522 J.

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

5. Step 5 — Evaluate:

   $$m = 143163\ \text{kg}$$

6. Step 6 — Check: returning m = 143,163 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

• 286,325 — kept a factor of two that cancels in the correct rearrangement.
• 71,581 — dropped that same factor in the other direction.
• 157,479 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Kinetics of a Rigid Body

A dynamics problem uses Work–energy theorem. Given mass (m) = 1,880 kg; initial speed (v1) = 3.5000 m/s; final speed (v2) = 34.0000 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) = 1,880 kg, initial speed (v1) = 3.5000 m/s, final speed (v2) = 34.0000 m/s.

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

5. Step 5 — Evaluate:

   $$W = 1075125\ \text{J}$$

6. Step 6 — Check: returning W = 1,075,125 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

• 2,150,250 — kept a factor of two that cancels in the correct rearrangement.
• 537,563 — dropped that same factor in the other direction.
• 1,182,638 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Kinetics of a Rigid Body

A dynamics problem uses Work–energy theorem. Given initial speed (v1) = 16.0000 m/s; final speed (v2) = 31.5000 m/s; work done (W) = 957,141 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) = 16.0000 m/s, final speed (v2) = 31.5000 m/s, work done (W) = 957,141 J.

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

5. Step 5 — Evaluate:

   $$m = 2600\ \text{kg}$$

6. Step 6 — Check: returning m = 2,600 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

• 5,200 — kept a factor of two that cancels in the correct rearrangement.
• 1,300 — dropped that same factor in the other direction.
• 2,860 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Kinetics of a Rigid Body

A dynamics problem uses Work–energy theorem. Given mass (m) = 1,200 kg; initial speed (v1) = 3.0000 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) = 1,200 kg, initial speed (v1) = 3.0000 m/s, final speed (v2) = 13.5000 m/s.

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

5. Step 5 — Evaluate:

   $$W = 103950\ \text{J}$$

6. Step 6 — Check: returning W = 103,950 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

• 207,900 — kept a factor of two that cancels in the correct rearrangement.
• 51,975 — dropped that same factor in the other direction.
• 114,345 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Kinetics of a Rigid Body

A dynamics problem uses Work–energy theorem. Given initial speed (v1) = 6.0000 m/s; final speed (v2) = 23.0000 m/s; work done (W) = 1,880,645 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) = 6.0000 m/s, final speed (v2) = 23.0000 m/s, work done (W) = 1,880,645 J.

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

5. Step 5 — Evaluate:

   $$m = 7629\ \text{kg}$$

6. Step 6 — Check: returning m = 7,629 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

• 15,259 — kept a factor of two that cancels in the correct rearrangement.
• 3,815 — dropped that same factor in the other direction.
• 8,392 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Kinetics of a Rigid Body