{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

HW5 Solution

# HW5 Solution - salazar(ggs332 HW05 TSOI(58160 This...

This preview shows pages 1–3. Sign up to view the full content.

salazar (ggs332) – HW05 – TSOI – (58160) 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A 0 . 4693 kg raindrop falls vertically at con- stant speed under the influence of gravity and air resistance. After the drop has fallen 156 m, what is the work done by gravity? The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 717 . 466 J. Explanation: The work is W = Force · Distance For the raindrop falling a distance h , the work done by the force of gravity m g is W g = m g h = (0 . 4693 kg) (9 . 8 m / s 2 ) (156 m) = 717 . 466 J . 002 (part 2 of 2) 10.0 points What is the work done by air resistance? Correct answer: - 717 . 466 J. Explanation: Since the raindrop remains at constant speed, the acceleration of the raindrop is zero, and Newton’s second law yields summationdisplay F y = F air - m g = 0 . It is evident that the force due to air resistance is equal in magnitude to the force due to gravity, but directed upward. The work done by air resistance can then be found, W air = F air h = - m g h = - (0 . 4693 kg) (9 . 8 m / s 2 ) × (156 m) = - 717 . 466 J . keywords: 003 10.0 points A cheerleader lifts his 60 . 5 kg partner straight up off the ground a distance of 0 . 845 m before releasing her. The acceleration of gravity is 9 . 8 m / s 2 . If he does this 12 times, how much work has he done? Correct answer: 6012 . 01 J. Explanation: The work done in lifting the cheerleader once is W 1 = m g h = (60 . 5 kg)(9 . 8 m / s 2 )(0 . 845 m) = 501 . 001 J . The work required to lift her n = 12 times is W = n W 1 = (12)(501 . 001 J) = 6012 . 01 J . 004 (part 1 of 3) 10.0 points A block of mass m is pushed a distance D up an inclined plane by a horizontal force F . The plane is inclined at an angle θ with respect to the horizontal. The block starts from rest and the coefficient of kinetic friction is μ k . m D μ k F θ If N is the normal force, what is the work done by friction? 1. W = - μ k ( N + m g cos θ ) D 2. W = - μ k ( N - m g cos θ ) D 3. W = + μ k N D 4. W = + μ k ( N - m g cos θ ) D

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
salazar (ggs332) – HW05 – TSOI – (58160) 2 5. W = + μ k ( N + m g cos θ ) D 6. W = 0 7. W = - μ k N D correct Explanation: The force of friction has a magnitude F friction = μ k N . Since it is in the direc- tion opposite to the motion, we get W friction = - F friction D = - μ k N D. 005 (part 2 of 3) 10.0 points What is the work done by the normal force N ? 1. W = ( m g cos θ + F sin θ - N ) D 2. W = N D sin θ 3. W = 0 correct 4. W = -N D 5. W = ( N - m g cos θ - F sin θ ) D 6. W = N D cos θ 7. W = ( N + m g cos θ + F sin θ ) D 8. W = N D Explanation: The normal force makes an angle of 90 with the displacement, so the work done by it is zero. 006 (part 3 of 3) 10.0 points What is the final speed of the block? 1. v = radicalbigg 2 m ( F cos θ - m g sin θ - μ k N ) D correct 2. v = radicalbigg 2 m ( F sin θ - μ k N ) D 3. v = radicalbigg 2 m ( F cos θ - μ k N ) D 4. v = radicalbigg 2 m ( F cos θ + m g sin θ ) D 5. v = radicalbigg 2 m ( F sin θ + μ k N ) D 6. v = radicalbigg 2 m ( F cos θ - m g sin θ + μ k N ) D 7. v = radicalbigg 2 m ( F cos θ + m g sin θ - μ k N ) D 8. v = radicalbigg 2 m ( F cos θ - m g sin θ ) D Explanation: The work done by gravity is W grav = m g D cos(90 + θ ) = - m g D sin θ .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 11

HW5 Solution - salazar(ggs332 HW05 TSOI(58160 This...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online