PHYS 303K HW 5

# PHYS 303K HW 5 - kodeih(njk359 – HW 5 – Coker –(56625...

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Unformatted text preview: kodeih (njk359) – HW 5 – Coker – (56625) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. This fairly difficult chapter (6) applies New- ton’s 2nd and 3rd Laws to examples with friction, with spring forces, and with circular motion at constant or variable speed. 001 10.0 points Two blocks are arranged at the ends of a massless string as shown in the figure. The system starts from rest. When the 3 . 69 kg mass has fallen through 0 . 42 m, its downward speed is 1 . 35 m / s. The acceleration of gravity is 9 . 8 m / s 2 . 3 . 69 kg 5 . 39 kg μ a What is the frictional force between the 5 . 39 kg mass and the table? Answer in units of N. 002 10.0 points A block of mass 7 kg rests on a plane inclined at an angle of 23 ◦ . The static coefficient of friction between the block and the plane is . 69. What is the frictional force on the block? 1. 43 . 5712 N 2. 63 . 1466 N 3. 26 . 8042 N 4. 18 . 4949 N 5. 16 . 767 N 003 10.0 points An object is held in place by friction on an inclined surface. The angle of inclination is increased until the object starts moving. If the surface is kept at this angle, the object 1. moves at uniform speed. 2. none of the other choices 3. speeds up. 4. slows down. 004 (part 1 of 2) 10.0 points A block of mass m is accelerated across a rough surface by a force of magnitude F that is exerted at an angle φ with the horizontal, as shown above. The frictional force on the block exerted by the surface has magnitude f . f F φ m What is the magnitude of the acceleration vectora of the block? 1. | vectora | = F cos φ − f m 2. | vectora | = F cos φ m 3. | vectora | = F − f m 4. | vectora | = F m 5. | vectora | = F sin φ − mg m 005 (part 2 of 2) 10.0 points Which of the following expressions for the coefficient of friction is correct? 1. μ = f mg 2. μ = f mg − F sin φ kodeih (njk359) – HW 5 – Coker – (56625) 2 3. μ = mg f 4. μ = mg − F cos φ f 5. μ = f mg − F cos φ 006 10.0 points A force vector F applied to a crate of mass M at an angle...
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PHYS 303K HW 5 - kodeih(njk359 – HW 5 – Coker –(56625...

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