Machinery's Handbook, 31st Edition
190 Work and Energy Example 1: A 12-inch cube of steel weighing 490 pounds is being moved on a horizontal conveyor belt at a speed of 6 miles per hour (8.8 feet per second). What is the kinetic energy of the cube? Solution: Since the block is not rotating, Formula (3a) for the kinetic energy of a body moving with pure translation applies: Kinetic Energy WV 2 2 g ------ 490 8.8 2 × 2 32.16 × ------------- 590 ft-lbs = = = A similar example using metric SI units is as follows: If a cube of mass 200 kg is being moved on a conveyor belt at a speed of 3 meters per second, what is the kinetic energy of the cube? It is: MV 2 1 2 1 200 3 900 Kinetic Energy joules 2 2 # # = = = Example 2: If the conveyor in Example 1 is brought to an abrupt stop, how long would it take for the steel block to come to a stop and how far along the belt would it slide before stopping if the coefficient of friction μ between the block and the conveyor belt is 0.2 and the block slides without tipping over? Solution: The only force acting to slow the motion of the block is the friction force be- tween the block and the belt. This force F is equal to the weight of the block W multiplied by the coefficient of friction; F = μ W = 0.2 3 490 = 98 lbs. The time required to bring the block to a stop can be determined from the impulse- momentum Formula (4c) on page 191 . R t × W g -- V f V o – ( ) –98 ( ) t 490 32.16 ------- 0 – 8.8 ( ) = = = t 490 8.8 × 98 32.16 × = ------------- 1.37 seconds = The distance the block slides before stopping can be determined by equating the kinetic energy of the block and the work done by friction in stopping it: Kinetic energy of block WV 2 2 g ------ Work done by friction F S ( × ) = 590 98 S × = S 590 98 = ----- = 6.0 feet If metric SI units are used, the calculation is as follows (for the cube of 200 kg mass): The friction force = μ multiplied by the weight Mg where g = approximately 9.81 m/s 2 . Thus, μ Mg = 0.2 3 200 g = 392.4 newtons. The time t required to bring the block to a stop is ( − 392.4) t = 200(0 − 3). Therefore, . . t 3924 200 3 1 53 seconds # = = The kinetic energy of the block is equal to the work done by friction, that is 392.4 3 S = 900 joules. Thus, the distance S which the block moves before stopping is . . S 3924 900 2 29 meters = = Force of a Blow.— A body that weighs W pounds and falls S feet from an initial position of rest is capable of doing WS foot-pounds of work. The work performed during its fall may, for example, be that necessary to drive a pile a distance d into the ground. Neglecting losses in the form of dissipated heat and strain energy, the work done in driving the pile is equal to the product of the impact force acting on the pile and the distance d which the
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