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HW05-solutions

# HW05-solutions - dang(ttd73 HW05 Tsoi(58020 This print-out...

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dang (ttd73) – HW05 – Tsoi – (58020) 1 This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (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 D 2. W = - μ k ( N - m g cos θ ) 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 D correct 7. W = 0 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. 002 (part 2 of 3) 10.0 points What is the work done by the normal force N ? 1. W = ( N - m g cos θ - F sin θ ) D 2. W = ( m g cos θ + F sin θ - N ) D 3. W = N D cos θ 4. W = N D sin θ 5. W = -N D 6. W = 0 correct 7. W = N D 8. W = ( N + m g cos θ + F sin θ ) D Explanation: The normal force makes an angle of 90 with the displacement, so the work done by it is zero. 003 (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 ) D 2. v = radicalbigg 2 m ( F sin θ + μ k N ) D 3. v = radicalbigg 2 m ( F sin θ - μ k N ) D 4. v = radicalbigg 2 m ( F cos θ - m g sin θ + μ k N ) D 5. v = radicalbigg 2 m ( F cos θ - m g sin θ - μ k N ) D correct 6. v = radicalbigg 2 m ( F cos θ - μ k N ) D 7. v = radicalbigg 2 m ( F cos θ - m g sin θ ) D 8. v = radicalbigg 2 m ( F cos θ + m g sin θ - μ k N ) D Explanation: The work done by gravity is W grav = m g D cos(90 + θ ) = - m g D sin θ . The work done by the force F is W F = F D cos θ .

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dang (ttd73) – HW05 – Tsoi – (58020) 2 From the work-energy theorem we know that W net = Δ K , W F + W grav + W friction = 1 2 m v 2 f . Thus v f = radicalbigg 2 m ( F cos θ - m g sin θ - μ k N ) D . 004 (part 1 of 2) 10.0 points A 500-N crate needs to be lifted 1 meter ver- tically in order to get it into the back of a pickup truck. What gives the crate a greater potential energy? 1. Either correct 2. Unable to determine 3. lift it straight up into the truck 4. slide it up a frictionless inclined plane Explanation: The change in height is the same for either method, so the new potential energies are the same. 005 (part 2 of 2) 10.0 points What is the advantage of using the inclined plane? 1. less distance 2. less force correct 3. less total energy 4. more power Explanation: An inclined plane can exchange an in- creased distance for less force. 006 (part 1 of 2) 10.0 points Consider a bungee cord of unstretched length L 0 = 35 m. When the cord is stretched to L > L 0 it behaves like a spring and obeys Hooke’s law with the spring constant k = 22 N / m. However, unlike a spring, the cord folds instead of becoming compressed when the distance between its ends is less than the unstretched length: For L < L 0 the cord has zero tension and zero elastic energy.
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HW05-solutions - dang(ttd73 HW05 Tsoi(58020 This print-out...

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