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1 A student raises one end of a board slowly, where a block lies. The block starts to move when the angle is 30º. The static friction coefficient between.

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Presentation on theme: "1 A student raises one end of a board slowly, where a block lies. The block starts to move when the angle is 30º. The static friction coefficient between."— Presentation transcript:

1 1 A student raises one end of a board slowly, where a block lies. The block starts to move when the angle is 30º. The static friction coefficient between the block and the board surface should be: 30º 1. a) b) c) d) e)None of above PHYSICS FOR ENGINEERS. 2006-2007. MID-TERM EXAM A1E. An object falls freely from rest on a planet without atmosphere where the gravity acceleration is 10 m s -2. Its velocity when it crashes against the ground is 5 m s -1, therefore its initial height was: 2. a) b) c) d) e)None of above 3. The angle formed by the vectorsandis: a) b) c) d) e)None of above “An object at rest stays at rest unless acted on by an external force”. This is: 4. a) b) c) d) e) None of above Newton’s third law Law of inertia Law of motion along a curved path The work-energy theorem A punctual mass follows a circular trajectory with constant speed. As for its acceleration, it is true that a) b) c) d) e) None of above The punctual mass is not accelerated Its tangential acceleration is positive The modulus of its normal acceleration is constant There is no normal acceleration 5. A particle whose mass is (1.00±0.01)  10 -2 kg is moving along a straight path at (1.00±0.10) m s -1. Its momentum is 6. a) b) c) d) e)None of above NAME: AGR FOR

2 2 A spring initially at rest, attached to a mass m, is stretched to distance x. As for the mass m, it is true that: 7. a) b) c) d) e) None of above The potential energy of m only varies in case the mass hangs vertically on the spring The potential energy of m varies if the mass hang vertically on the spring, but even if the spring and the mass lie on a horizontal surface. The potential energy of m never changes, only its kinetic energy undergoes some variation. Neither the potential energy of m nor its kinetic energy undergo any variation. Two blocks of masses m 1 and m 2 lie on a horizontal table. On m 1 we apply a horizontal force F 0 as shown in the picture. The kinetic friction coefficients for both masses are  1 y  2, respectively. Using the numerical values given below, answer the following questions: m1m1 m2m2 F0F0 Assuming the force F 0 is big enough to move the set of two blocks, find its acceleration. b) c) Find the force exerted by the first block (m 1 ) on the second one (m 2 ) and the force exerted by the second one on the first one.  1 = 0.075  2 = 0.040 m 1 = 8 kgF 0 = 2,50 kpm 2 = 6 kg Separately draw the free body diagram for the set of two blocks, for m 1 and for m 2. a) PROBLEM GRADING: PROBLEM: 6 POINTS QUESTIONS: 4 POINTS EACH CORRECT ANSWER: +0.500 EACH WRONG ANSWER: -0.125 A harmonic oscillator obeys the equationwhere every quantity is given in S.I. units. The frequency and the period are: 8. 4.5 Hz and 0.22 sa)0.5 Hz and 2.0 s b) 6.3 Hz and 0.16 sc)5 Hz and 9.90 s d) None of above e)

3 3 A student raises one end of a board slowly, where a block lies. The block starts to move when the angle is 30º. The static friction coefficient between the block and the board surface should be: 30º 1. a) b) c) d) e)None of above An object falls freely from rest on a planet without atmosphere where the gravity acceleration is 10 m s -2. Its velocity when it crashes against the ground is 5 m s -1, therefore its initial height was: 2. a) b) c) d) e)None of above 3. The angle formed by the vectorsandis: a) b) c) d) e)None of above “An object at rest stays at rest unless acted on by an external force”. This is: 4. a) b) c) d) e) None of above Newton’s third law Law of inertia Law of motion along a curved path The work-energy theorem A punctual mass follows a circular trajectory with constant speed. As for its acceleration, it is true that a) b) c) d) e) None of above The punctual mass is not accelerated Its tangential acceleration is positive The modulus of its normal acceleration is constant There is no normal acceleration 5. A particle whose mass is (1.00±0.01)  10 -2 kg is moving along a straight path at (1.00±0.10) m s -1. Its momentum is 6. a) b) c) d) e)None of above PHYSICS FOR ENGINEERS. 2006-2007. MID-TERM EXAM A1. SOLUTION

4 4 A spring initially at rest, attached to a mass m, is stretched to distance x. As for the mass m, it is true that: 7. a) b) c) d) e) None of above The potential energy of m only varies in case the mass hangs vertically on the spring The potential energy of m varies if the mass hang vertically on the spring, but even if the spring and the mass lie on a horizontal surface. The potential energy of m never changes, only its kinetic energy undergoes some variation. Neither the potential energy of m nor its kinetic energy undergo any variation. Two blocks of masses m 1 and m 2 lie on a horizontal table. On m 1 we apply a horizontal force F 0 as shown in the picture. The kinetic friction coefficients for both masses are  1 y  2, respectively. Using the numerical values given below, answer the following questions: m1m1 m2m2 F0F0 Assuming the force F 0 is big enough to move the set of two blocks, find its acceleration. b) c) Find the force exerted by the first block (m 1 ) on the second one (m 2 ) and the force exerted by the second one on the first one.  1 = 0.075  2 = 0.040 m 1 = 8 kgF 0 = 2,50 kpm 2 = 6 kg Separately draw the free body diagram for the set of two blocks, for m 1 and for m 2. a) PROBLEM GRADING: PROBLEM: 6 POINTS QUESTIONS: 4 POINTS EACH CORRECT ANSWER: +0.500 EACH WRONG ANSWER: -0.125 A harmonic oscillator obeys the equationwhere every quantity is given in S.I. units. The frequency and the period are: 8. 4.5 Hz and 0.22 sa)0.5 Hz and 2.0 s b) 6.3 Hz and 0.16 sc)5 Hz and 9.90 s d) None of above SOLUTION(CONTINUED) e)

5 5 m1m1 m2m2 F0F0 F 21 F R1 F0F0 F 12 F R2 N2N2 F R1 F R2 N1N1 N2N2 Newton’s 2 nd law Friction forces Newton’s 2 nd law  1 = 0.075  2 = 0.040 m 1 = 8 kgF 0 = 2,50 kpm 2 = 6 kg Numerical result a = 1.162 m/s 2 Newton’s 2 nd law F 12 and F 21 have to be equal (action and reaction) N1N1 SOLUTION(CONTINUED)


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