{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# HW 5 - Morgan Mena Homework 5 Due 11:00 pm Inst Opyrchal H...

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

Morgan, Mena – Homework 5 – Due: Oct 10 2007, 11:00 pm – Inst: Opyrchal, H 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points A hockey puck is given an initial speed of 24 m / s on a frozen pond. The puck remains on the ice and slides 118 m before coming to rest. The acceleration of gravity is 9 . 8 m / s 2 . What is the coefficient of friction between the puck and the ice? Correct answer: 0 . 249049 . Explanation: Basic Concept: Newton’s Second Law: F net = ma Solution: From kinematics, 0 = v 2 0 + 2 ax a = - v 2 0 2 x since the puck comes to a rest. Consider the free body diagram for the situation: μ N N mg Vertically, the puck is in equilibrium, so F net = N - mg = 0 ⇒ N = mg Horizontally, after the puck is given its initial velocity, the friction force f k = μ N = μmg is the only unbalanced force acting on it. Thus, f net = ma = 0 - f k = - μmg μ = - a g = v 2 2 gx = (24 m / s) 2 2(9 . 8 m / s 2 )(118 m) = 0 . 249049 keywords: 002 (part 1 of 1) 10 points A 4 . 9 kg block slides down an inclined plane that makes an angle of 25 with the horizontal. Starting from rest, the block slides a distance of 2 . 2 m in 4 . 9 s. The acceleration of gravity is 9 . 81 m / s 2 . Find the coefficient of kinetic friction be- tween the block and plane. Correct answer: 0 . 445696 . Explanation: Let : m = 4 . 9 kg , θ = 25 , Δ x = 2 . 2 m , and Δ t = 4 . 9 s . 4 . 9 kg μ 25 Consider the forces acting on the block as it slides down the incline: ~ N ~ f k m~g From kinematics, since v 0 = 0, Δ x = v 0 Δ t + 1 2 a t ) 2 = 1 2 a t ) 2 a = 2 Δ x t ) 2 = 2 (2 . 2 m) (4 . 9 s) 2 = 0 . 183257 m / s 2 .

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

View Full Document
Morgan, Mena – Homework 5 – Due: Oct 10 2007, 11:00 pm – Inst: Opyrchal, H 2 Applying X ~ F = m~a to the block, X F y = N - m g cos θ = 0 m g cos θ = N and X F x = m g sin θ - f k = m a m g sin θ - μ k N = m a m g sin θ - μ k m g cos θ = m a . Thus μ k = g sin θ - a g cos θ = (9 . 81 m / s 2 ) sin 25 - (0 . 183257 m / s 2 ) (9 . 81 m / s 2 ) cos 25 = 0 . 445696 . keywords: 003 (part 1 of 2) 10 points A 8 kg block rests on a horizontal table, at- tached to a 7 kg block by a light string as shown in the figure. The acceleration of gravity is 9 . 81 m / s 2 . 8 kg 7 kg ~ T What is the minimum coefficient of static friction such that the objects remain at rest?
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 7

HW 5 - Morgan Mena Homework 5 Due 11:00 pm Inst Opyrchal H...

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

View Full Document
Ask a homework question - tutors are online