A 5079 lb vehicle travelling 70 ft/s must stop in 247 ft. What constant resisting force is required?

Start with the thinking

• Work done by the resisting force equals the change in kinetic energy.
• Mass, not weight, appears in kinetic energy.

Step-by-step solution

1. Mass

   $$m = 5079/32.2 = 157.7 slugs$$

2. Kinetic energy

   $$KE = ½mv^{2}$$

3. Substituting — KE = 0.5(157.7)(70²) = 386,446 ft·lb

4. Work–energy — F·d = KE

5. Substituting

   $$F = 386,446/247 = 1,565 lb$$

Why the other options are there

• 50,379 lb (weight used as mass)
• 386,446 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Power and Efficiency

A 3169 lb vehicle travelling 60 ft/s must stop in 245 ft. What constant resisting force is required?

Start with the thinking

• Work done by the resisting force equals the change in kinetic energy.
• Mass, not weight, appears in kinetic energy.

Step-by-step solution

1. Mass

   $$m = 3169/32.2 = 98.42 slugs$$

2. Kinetic energy

   $$KE = ½mv^{2}$$

3. Substituting — KE = 0.5(98.42)(60²) = 177,149 ft·lb

4. Work–energy — F·d = KE

5. Substituting

   $$F = 177,149/245 = 723.1 lb$$

Why the other options are there

• 23,282 lb (weight used as mass)
• 177,149 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Power and Efficiency

A 6283 lb vehicle travelling 48 ft/s must stop in 121 ft. What constant resisting force is required?

Start with the thinking

• Work done by the resisting force equals the change in kinetic energy.
• Mass, not weight, appears in kinetic energy.

Step-by-step solution

1. Mass

   $$m = 6283/32.2 = 195.1 slugs$$

2. Kinetic energy

   $$KE = ½mv^{2}$$

3. Substituting — KE = 0.5(195.1)(48²) = 224,783 ft·lb

4. Work–energy — F·d = KE

5. Substituting

   $$F = 224,783/121 = 1,858 lb$$

Why the other options are there

• 59,818 lb (weight used as mass)
• 224,783 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Power and Efficiency

A 3535 lb vehicle travelling 65 ft/s must stop in 214 ft. What constant resisting force is required?

Start with the thinking

• Work done by the resisting force equals the change in kinetic energy.
• Mass, not weight, appears in kinetic energy.

Step-by-step solution

1. Mass

   $$m = 3535/32.2 = 109.8 slugs$$

2. Kinetic energy

   $$KE = ½mv^{2}$$

3. Substituting — KE = 0.5(109.8)(65²) = 231,916 ft·lb

4. Work–energy — F·d = KE

5. Substituting

   $$F = 231,916/214 = 1,084 lb$$

Why the other options are there

• 34,896 lb (weight used as mass)
• 231,916 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Power and Efficiency

A 3564 lb vehicle travelling 53 ft/s must stop in 179 ft. What constant resisting force is required?

Start with the thinking

• Work done by the resisting force equals the change in kinetic energy.
• Mass, not weight, appears in kinetic energy.

Step-by-step solution

1. Mass

   $$m = 3564/32.2 = 110.7 slugs$$

2. Kinetic energy

   $$KE = ½mv^{2}$$

3. Substituting — KE = 0.5(110.7)(53²) = 155,455 ft·lb

4. Work–energy — F·d = KE

5. Substituting

   $$F = 155,455/179 = 868.5 lb$$

Why the other options are there

• 27,964 lb (weight used as mass)
• 155,455 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Power and Efficiency

A 2645 lb vehicle travelling 47 ft/s must stop in 177 ft. What constant resisting force is required?

Start with the thinking

• Work done by the resisting force equals the change in kinetic energy.
• Mass, not weight, appears in kinetic energy.

Step-by-step solution

1. Mass

   $$m = 2645/32.2 = 82.14 slugs$$

2. Kinetic energy

   $$KE = ½mv^{2}$$

3. Substituting — KE = 0.5(82.14)(47²) = 90,727 ft·lb

4. Work–energy — F·d = KE

5. Substituting

   $$F = 90,727/177 = 512.6 lb$$

Why the other options are there

• 16,505 lb (weight used as mass)
• 90,727 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Power and Efficiency

A 5576 lb vehicle travelling 45 ft/s must stop in 125 ft. What constant resisting force is required?

Start with the thinking

• Work done by the resisting force equals the change in kinetic energy.
• Mass, not weight, appears in kinetic energy.

Step-by-step solution

1. Mass

   $$m = 5576/32.2 = 173.2 slugs$$

2. Kinetic energy

   $$KE = ½mv^{2}$$

3. Substituting — KE = 0.5(173.2)(45²) = 175,332 ft·lb

4. Work–energy — F·d = KE

5. Substituting

   $$F = 175,332/125 = 1,403 lb$$

Why the other options are there

• 45,166 lb (weight used as mass)
• 175,332 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Power and Efficiency

A 3906 lb vehicle travelling 47 ft/s must stop in 205 ft. What constant resisting force is required?

Start with the thinking

• Work done by the resisting force equals the change in kinetic energy.
• Mass, not weight, appears in kinetic energy.

Step-by-step solution

1. Mass

   $$m = 3906/32.2 = 121.3 slugs$$

2. Kinetic energy

   $$KE = ½mv^{2}$$

3. Substituting — KE = 0.5(121.3)(47²) = 133,981 ft·lb

4. Work–energy — F·d = KE

5. Substituting

   $$F = 133,981/205 = 653.6 lb$$

Why the other options are there

• 21,045 lb (weight used as mass)
• 133,981 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Power and Efficiency

A 6985 lb vehicle travelling 34 ft/s must stop in 105 ft. What constant resisting force is required?

Start with the thinking

• Work done by the resisting force equals the change in kinetic energy.
• Mass, not weight, appears in kinetic energy.

Step-by-step solution

1. Mass

   $$m = 6985/32.2 = 216.9 slugs$$

2. Kinetic energy

   $$KE = ½mv^{2}$$

3. Substituting — KE = 0.5(216.9)(34²) = 125,383 ft·lb

4. Work–energy — F·d = KE

5. Substituting

   $$F = 125,383/105 = 1,194 lb$$

Why the other options are there

• 38,451 lb (weight used as mass)
• 125,383 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Power and Efficiency

A 4867 lb vehicle travelling 60 ft/s must stop in 210 ft. What constant resisting force is required?

Start with the thinking

• Work done by the resisting force equals the change in kinetic energy.
• Mass, not weight, appears in kinetic energy.

Step-by-step solution

1. Mass

   $$m = 4867/32.2 = 151.1 slugs$$

2. Kinetic energy

   $$KE = ½mv^{2}$$

3. Substituting — KE = 0.5(151.1)(60²) = 272,068 ft·lb

4. Work–energy — F·d = KE

5. Substituting

   $$F = 272,068/210 = 1,296 lb$$

Why the other options are there

• 41,717 lb (weight used as mass)
• 272,068 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Power and Efficiency