Old_Midterm-2B - practicework 02 – MENDIOLA NICHOLE L –...

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Unformatted text preview: practicework 02 – MENDIOLA, NICHOLE L – Due: Mar 6 2006, 4:00 am 1 Version number encoded for clicker entry: V1:1, V2:1, V3:4, V4:5, V5:3. Question 1 Part 1 of 3. 10 points. The suspended 2 . 7 kg mass on the right is moving up, the 2 . 3 kg mass slides down the ramp, and the suspended 8 . 3 kg mass on the left is moving down. The coefficient of friction between the block and the ramp is 0 . 12 . The acceleration of gravity is 9 . 8 m / s 2 . The pulleys are massless and frictionless. 2 . 3 k g μ = . 1 2 31 ◦ 8 . 3 kg 2 . 7 kg What is the acceleration of the three block system? Answer in units of m / s 2 . Question 2 Part 2 of 3. 10 points. What is the tension in the cord connected to the 2 . 7 kg block? Answer in units of N. Question 3 Part 3 of 3. 10 points. What is the tension in the cord connected to the 8 . 3 kg block? Answer in units of N. Question 4 Part 1 of 3. 10 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 D 2. W = + μ k N D 3. W = + μ k ( N - m g cos θ ) D 4. W =- μ k ( N - m g cos θ ) D 5. W =- μ k ( N + m g cos θ ) D 6. W = + μ k ( N + m g cos θ ) D 7. W = 0 Question 5 Part 2 of 3. 10 points. What is the work done by the normal force N ? 1. W = 0 2. W =-N D 3. W = N D 4. W = N D cos θ 5. W = ( N + m g cos θ + F sin θ ) D 6. W = ( m g cos θ + F sin θ- N ) D 7. W = ( N - m g cos θ- F sin θ ) D practicework 02 – MENDIOLA, NICHOLE L – Due: Mar 6 2006, 4:00 am 2 8. W = N D sin θ Question 6 Part 3 of 3. 10 points. What is the final speed of the block? 1. v = r 2 m ( F cos θ- m g sin θ + μ k N ) D 2. v = r 2 m ( F sin θ- μ k N ) D 3. v = r 2 m ( F cos θ + m g sin θ ) D 4. v = r 2 m ( F sin θ + μ k N ) D 5. v = r 2 m ( F cos θ- μ k N ) D 6. v = r 2 m ( F cos θ- m g sin θ ) D 7. v = r 2 m ( F cos θ + m g sin θ- μ k N ) D 8. v = r 2 m ( F cos θ- m g sin θ- μ k N ) D Question 7 Part 1 of 2. 10 points. Given: The coefficient of static friction be- tween the person and the wall is 0 . 62 , the mass of the person is 66 kg , the radius of the cylinder is 7 . 5 m , and g = 9 . 8 m / s 2 . A barrel of fun consists of a large vertical cylinder that spins about the vertical axis. When it spins fast enough, any person inside will be held up against the wall. ω 7 . 5 m Find ω c , the critical angular speed below which a person will slide down the wall of the cylinder. Answer in units of rad / s. Question 8 Part 2 of 2. 10 points. Compare the following two cases where for each case the angular speed is greater than the critical value as defined in Part 1. For the first case, denote its angular speed and frictional force by ω 1 and f 1 , respectively. For the second case, its angular speed is ω 2 = 2 ω 1 ....
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Old_Midterm-2B - practicework 02 – MENDIOLA NICHOLE L –...

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