cleared of snow‚ the friction between the ground and the skis brings our hero to a halt at point D‚ located at a distance d = 12 m from point C. The descent can be considered frictionless. Take the potential energy to be zero at the bottom of the slope. (a) What is the mechanical energy of James Bond at points A and D? (b) Determine the speed of Bond at position B abd C. (c) What is the work done by friction in the parking lot? (d) Find the magnitude of the average friction force. SOLUTION : (a)
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a critical depth of penetration. Some sensitivity analyses have been performed to evaluate the effect of soil‚ pile and hammer properties on the level of vibrations. The results show that increase in pile diameter‚ hammer impact force‚ soil–pile friction and reduction in soil elastic modulus can increase the peak particle velocity. Notation D d E e L Lmin p r VP VR VS Æ â ˜t ¨ ì í r ô ôcrit ö pile
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Part B Now‚ suppose that Zak’s younger cousin‚ Greta‚ sees him sliding and takes off her shoes so that she can slide as well (assume her socks have the same coefficient of kinetic friction as Zak’s). Instead of getting a running start‚ she asks Zak to give her a push. So‚ Zak pushes her with a force of 125 \rm N over a distance of 1.00 \rm m. If her mass is 20.0 \rm kg‚ what distance d_2 does she slide after Zak’s push ends? Remember that the frictional force acts on Greta during Zak’s push and
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a 40 teeth gear set. To start the experiment‚ the effort pulley and the load drum were loaded with appropriate weights and wound. After the weights were loaded‚ a small downward pull was given to the effort hanger in order to overcome the static friction. It was advised that this manual downward pull should be given by the same person and the force should be the same throughout the whole experiment since it could affect the experiment results adversely if done otherwise. And it was also advised that
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UNIVERSITI TUN HUSSEIN ONN MALAYSIA FINAL EXAMINATION SEMESTER II SESSION 2011/2012 COURSE NAME COURSE CODE PROGRAMME EXAMINATION DATE DURATION INSTRUCTION : : : : : : MECHANICS OF MACHINES BDA20303 / BDA2033 2 BDD JUNE 2012 3 HOURS PART A: ANSWER ALL QUESTIONS PART B: ANSWER ONE (1) QUESTION ONLY THIS PAPER CONTAINS SEVEN (7) PAGES BDA20303 / BDA2033 PART A (COMPULSORY) Q1 (a) Indicate ‘True’ or ‘False’ for the following statements: The radial distance between pitch circle
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Prac Report Problem: How does the increase mass affect acceleration and the force of the accelerating object? Purpose: The purpose of the practical is to find how mass affects acceleration and how it affects also the force of the accelerating body. To do this we are going to do the ticker tape experiment where an accelerating body pulls a tape through a consistent 50 dot per second ticker timer. The acceleration body in this experiment will be a small trolley pulled by a string that is pulled
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experience an upward force called air resistance (drag). RAF parachutist Objects with large surface areas‚ such as parachutes or shuttlecocks fall more slowly because they experience more air resistance. Frictional forces such as air resistance‚ friction and drag act against the direction of motion‚ so tend to slow the object down. This fact is put to good use in the design of the parachute and shuttlecock. The size of frictional forces can be reduced bystreamlining the object or lubricating any
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Experiment 9: Maxwell’s Wheel Introduction: The second lab performed on 2/1/12 involved two investigations concerning Maxwell’s wheel. Maxwell’s wheel is an apparatus that consists of a large disk with a long axle. The disk then bound to a support hanging from above with strings attached to each end of the axle. Maxwell’s wheel is considered to be an important apparatus to investigate physical phenomenon’s because it its ability to combine straight line motion and rotation of a rigid body
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box is being pulled with a force of 12 N (See Figure 1). The position-time graph of the box is given in Figure 2. a. Sketch a free body diagram of the forces acting on the box. (2 marks) b. Determine the coefficient of friction between box and
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about 30 meters from edge when it starts skidding through dirt and sand. Newton second law says Fnet = Ffriction = µmg = ma where the acceleration of the car is completely due to the friction force. M is the mass of the car‚ g is equal to the acceleration due to gravity (9.8m/s2) ‚ µ is the coefficient of sliding friction between sand and tires (0.5 at most)‚ and a is the acceleration of the car. Solving for a we get: a = µg = (0.5)(9.8m/s2) = 4.9 m/s2If we assume a relatively constant acceleration
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