A 3,500 lb pile hammer falls freely 4.0 ft. What is its kinetic energy and impact velocity?

Start with the thinking

• Energy conservation: potential energy converts fully to kinetic in free fall.
• Mass in slugs equals W/g.

Step-by-step solution

1. Energy — KE = W h = 3,500(4.0) = 14,000 ft·lb

2. Mass

   $$m = W/g = 3,500/32.2 = 108.7 slug$$

3. Velocity relation

   $$KE = ½ m v^{2}$$

4. Rearrange

   $$v = \sqrt(2KE/m) = \sqrt(28,000/108.7)$$

5. Result

   $$v = 16.0 ft/s$$

Why the other options are there

• v = 8.0 ft/s (factor of 2 omitted)
• KE = 435 ft·lb (weight divided by g twice)

Reference: FE Reference Handbook — Dynamics — Work and energy

A dynamics problem uses Work–energy theorem. Given mass (m) = 1,550 kg; initial speed (v1) = 18.0000 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) = 1,550 kg, initial speed (v1) = 18.0000 m/s, final speed (v2) = 39.5000 m/s.

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

5. Step 5 — Evaluate:

   $$W = 958094\ \text{J}$$

6. Step 6 — Check: returning W = 958,094 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

• 1,916,188 — kept a factor of two that cancels in the correct rearrangement.
• 479,047 — dropped that same factor in the other direction.
• 1,053,903 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Kinetic Energy

A dynamics problem uses Work–energy theorem. Given initial speed (v1) = 14.5000 m/s; final speed (v2) = 29.0000 m/s; work done (W) = 1,828,922 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) = 14.5000 m/s, final speed (v2) = 29.0000 m/s, work done (W) = 1,828,922 J.

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

5. Step 5 — Evaluate:

   $$m = 5799\ \text{kg}$$

6. Step 6 — Check: returning m = 5,799 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

• 11,598 — kept a factor of two that cancels in the correct rearrangement.
• 2,900 — dropped that same factor in the other direction.
• 6,379 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Kinetic Energy

A dynamics problem uses Work–energy theorem. Given mass (m) = 1,980 kg; initial speed (v1) = 12.5000 m/s; final speed (v2) = 9.0000 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) = 1,980 kg, initial speed (v1) = 12.5000 m/s, final speed (v2) = 9.0000 m/s.

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

5. Step 5 — Evaluate:

   $$W = -74498\ \text{J}$$

6. Step 6 — Check: returning W = -74,498 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

• -148,995 — kept a factor of two that cancels in the correct rearrangement.
• -37,249 — dropped that same factor in the other direction.
• -81,947 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Kinetic Energy

A dynamics problem uses Work–energy theorem. Given initial speed (v1) = 11.5000 m/s; final speed (v2) = 27.5000 m/s; work done (W) = 251,587 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) = 11.5000 m/s, final speed (v2) = 27.5000 m/s, work done (W) = 251,587 J.

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

5. Step 5 — Evaluate:

   $$m = 806.4\ \text{kg}$$

6. Step 6 — Check: returning m = 806.4 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

• 1,613 — kept a factor of two that cancels in the correct rearrangement.
• 403.2 — dropped that same factor in the other direction.
• 887.0 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Kinetic Energy

A dynamics problem uses Work–energy theorem. Given mass (m) = 1,330 kg; initial speed (v1) = 16.5000 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) = 1,330 kg, initial speed (v1) = 16.5000 m/s, final speed (v2) = 30.0000 m/s.

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

5. Step 5 — Evaluate:

   $$W = 417454\ \text{J}$$

6. Step 6 — Check: returning W = 417,454 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

• 834,908 — kept a factor of two that cancels in the correct rearrangement.
• 208,727 — dropped that same factor in the other direction.
• 459,199 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Kinetic Energy

A dynamics problem uses Work–energy theorem. Given initial speed (v1) = 12.0000 m/s; final speed (v2) = 6.5000 m/s; work done (W) = 706,226 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) = 12.0000 m/s, final speed (v2) = 6.5000 m/s, work done (W) = 706,226 J.

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

5. Step 5 — Evaluate:

   $$m = -13882\ \text{kg}$$

6. Step 6 — Check: returning m = -13,882 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

• -27,763 — kept a factor of two that cancels in the correct rearrangement.
• -6,941 — dropped that same factor in the other direction.
• -15,270 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Kinetic Energy

A dynamics problem uses Work–energy theorem. Given mass (m) = 930.0 kg; initial speed (v1) = 2.5000 m/s; final speed (v2) = 18.0000 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) = 930.0 kg, initial speed (v1) = 2.5000 m/s, final speed (v2) = 18.0000 m/s.

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

5. Step 5 — Evaluate:

   $$W = 147754\ \text{J}$$

6. Step 6 — Check: returning W = 147,754 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

• 295,508 — kept a factor of two that cancels in the correct rearrangement.
• 73,877 — dropped that same factor in the other direction.
• 162,529 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Kinetic Energy

A dynamics problem uses Work–energy theorem. Given initial speed (v1) = 10.0000 m/s; final speed (v2) = 8.5000 m/s; work done (W) = 1,146,758 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) = 8.5000 m/s, work done (W) = 1,146,758 J.

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

5. Step 5 — Evaluate:

   $$m = -82649\ \text{kg}$$

6. Step 6 — Check: returning m = -82,649 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

• -165,298 — kept a factor of two that cancels in the correct rearrangement.
• -41,325 — dropped that same factor in the other direction.
• -90,914 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Kinetic Energy

A dynamics problem uses Work–energy theorem. Given mass (m) = 2,460 kg; initial speed (v1) = 19.0000 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) = 2,460 kg, initial speed (v1) = 19.0000 m/s, final speed (v2) = 14.0000 m/s.

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

5. Step 5 — Evaluate:

   $$W = -202950\ \text{J}$$

6. Step 6 — Check: returning W = -202,950 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

• -405,900 — kept a factor of two that cancels in the correct rearrangement.
• -101,475 — dropped that same factor in the other direction.
• -223,245 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Kinetic Energy