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) = 980.0 kg; initial speed (v1) = 1.5000 m/s; final speed (v2) = 28.0000 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) = 980.0 kg, initial speed (v1) = 1.5000 m/s, final speed (v2) = 28.0000 m/s.

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

5. Step 5 — Evaluate:

   $$W = 383058\ \text{J}$$

6. Step 6 — Check: returning W = 383,058 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

• 766,115 — kept a factor of two that cancels in the correct rearrangement.
• 191,529 — dropped that same factor in the other direction.
• 421,363 — 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) = 33.0000 m/s; work done (W) = 1,890,653 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) = 33.0000 m/s, work done (W) = 1,890,653 J.

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

5. Step 5 — Evaluate:

   $$m = 3952\ \text{kg}$$

6. Step 6 — Check: returning m = 3,952 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

• 7,904 — kept a factor of two that cancels in the correct rearrangement.
• 1,976 — dropped that same factor in the other direction.
• 4,347 — 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) = 2,410 kg; initial speed (v1) = 19.0000 m/s; final speed (v2) = 39.0000 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) = 2,410 kg, initial speed (v1) = 19.0000 m/s, final speed (v2) = 39.0000 m/s.

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

5. Step 5 — Evaluate:

   $$W = 1397800\ \text{J}$$

6. Step 6 — Check: returning W = 1,397,800 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

• 2,795,600 — kept a factor of two that cancels in the correct rearrangement.
• 698,900 — dropped that same factor in the other direction.
• 1,537,580 — 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) = 4.0000 m/s; final speed (v2) = 31.5000 m/s; work done (W) = 1,323,002 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) = 4.0000 m/s, final speed (v2) = 31.5000 m/s, work done (W) = 1,323,002 J.

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

5. Step 5 — Evaluate:

   $$m = 2710\ \text{kg}$$

6. Step 6 — Check: returning m = 2,710 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

• 5,421 — kept a factor of two that cancels in the correct rearrangement.
• 1,355 — dropped that same factor in the other direction.
• 2,981 — 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) = 2,800 kg; initial speed (v1) = 13.5000 m/s; final speed (v2) = 22.0000 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) = 2,800 kg, initial speed (v1) = 13.5000 m/s, final speed (v2) = 22.0000 m/s.

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

5. Step 5 — Evaluate:

   $$W = 422450\ \text{J}$$

6. Step 6 — Check: returning W = 422,450 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

• 844,900 — kept a factor of two that cancels in the correct rearrangement.
• 211,225 — dropped that same factor in the other direction.
• 464,695 — 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) = 2.5000 m/s; final speed (v2) = 18.5000 m/s; work done (W) = 65,624 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) = 2.5000 m/s, final speed (v2) = 18.5000 m/s, work done (W) = 65,624 J.

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

5. Step 5 — Evaluate:

   $$m = 390.6\ \text{kg}$$

6. Step 6 — Check: returning m = 390.6 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

• 781.2 — kept a factor of two that cancels in the correct rearrangement.
• 195.3 — dropped that same factor in the other direction.
• 429.7 — 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) = 2,230 kg; initial speed (v1) = 9.5000 m/s; final speed (v2) = 12.0000 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) = 2,230 kg, initial speed (v1) = 9.5000 m/s, final speed (v2) = 12.0000 m/s.

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

5. Step 5 — Evaluate:

   $$W = 59931\ \text{J}$$

6. Step 6 — Check: returning W = 59,931 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

• 119,863 — kept a factor of two that cancels in the correct rearrangement.
• 29,966 — dropped that same factor in the other direction.
• 65,924 — 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) = 16.5000 m/s; final speed (v2) = 15.5000 m/s; work done (W) = 1,889,207 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) = 15.5000 m/s, work done (W) = 1,889,207 J.

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

5. Step 5 — Evaluate:

   $$m = -118075\ \text{kg}$$

6. Step 6 — Check: returning m = -118,075 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

• -236,151 — kept a factor of two that cancels in the correct rearrangement.
• -59,038 — dropped that same factor in the other direction.
• -129,883 — 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) = 810.0 kg; initial speed (v1) = 9.0000 m/s; final speed (v2) = 8.5000 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) = 810.0 kg, initial speed (v1) = 9.0000 m/s, final speed (v2) = 8.5000 m/s.

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

5. Step 5 — Evaluate:

   $$W = -3544\ \text{J}$$

6. Step 6 — Check: returning W = -3,544 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

• -7,088 — kept a factor of two that cancels in the correct rearrangement.
• -1,772 — dropped that same factor in the other direction.
• -3,898 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Kinetic Energy