OH-06-solutions

# OH-06-solutions - alsalhi (raa2359) – OH-06 – bradley...

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

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: alsalhi (raa2359) – OH-06 – bradley – (19102) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. This is Online Homework No. 6 (OH-06). All answers should be submitted online be- fore 2300 on Tuesday 8 March (UAE time). Computers and network connections can give trouble, so do not wait until the last minute! 001 10.0 points A 4 kg object attached to a horizontal string moves with constant speed in a circle on a frictionless horizontal surface. The kinetic energy of the object is 17 J and the tension in the string is 306 N. Find the radius of the circle. Correct answer: 0 . 111111 m. Explanation: Let : m = 4 kg , K = 17 J , and T = 306 N . T N mg r ω The kinetic energy is K = 1 2 mv 2 . The centripital force is supplied by the string: T = F c = mv 2 r r = mv 2 T = 2 K T = 2 (17 J) 306 N = . 111111 m . keywords: 002 10.0 points A 2 . 25 kg block is pushed 1 . 66 m up a vertical wall with constant speed by a constant force of magnitude F applied at an angle of 64 . 5 ◦ with the horizontal. The acceleration of gravity is 9 . 8 m / s 2 . 2 . 25 kg F 6 4 . 5 ◦ If the coefficient of kinetic friction between the block and wall is 0 . 669, find the work done by F . Correct answer: 53 . 7565 J. Explanation: Given : m = 2 . 25 kg , μ = 0 . 669 , θ = 64 . 5 ◦ , and Δ y = 1 . 66 m . F 6 4 . 5 ◦ v mg f k N The block is in equilibrium horizontally, so summationdisplay F x = F cos θ-N = 0 , so that N = F cos θ alsalhi (raa2359) – OH-06 – bradley – (19102) 2 Since the block moves with constant velocity, summationdisplay F y = F sin θ- mg- f k = 0 F sin θ- mg- μ F cos θ = 0 F (sin θ- μ cos θ ) = mg , so F = mg sin θ- μ cos θ = (2 . 25 kg) (9 . 8 m / s 2 ) sin 64 . 5 ◦- . 669 cos 64 . 5 ◦ = 35 . 8785 N . Thus W F = ( F sin θ ) (Δ y ) = (35 . 8785 N) (sin64 . 5 ◦ )(1 . 66 m) = 53 . 7565 J , since 1 J = 1 kg · m 2 / s 2 . 003 10.0 points Starting from rest at a height equal to the radius of the circular track, a block of mass 23 kg slides down a quarter circular track under the influence of gravity with friction present (of coefficient μ ). The radius of the track is 34 m. The acceleration of gravity is 9 . 8 m / s 2 . 34 m 23 kg θ If the kinetic energy of the block at the bottom of the track is 5000 J, what is the work done against friction? Correct answer: 2663 . 6 J. Explanation: W = W f + K W f = mg R- K = (23 kg) (9 . 8 m / s 2 ) (34 m)- (5000 J) = 2663 . 6 J . 004 (part 1 of 2) 10.0 points A 5 . 07 × 10 − 5 kg raindrop falls vertically at constant speed under the influence of gravity and air resistance....
View Full Document

## This note was uploaded on 05/03/2011 for the course PHYS 191 taught by Professor Aboc during the Spring '10 term at The Petroleum Institute.

### Page1 / 7

OH-06-solutions - alsalhi (raa2359) – OH-06 – bradley...

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

View Full Document
Ask a homework question - tutors are online