Question

PHYSICS HOMEWORK # 26 NEWTON'S LAWS FIRST LAW & FRICTION 24. Two masses are sitting on a horizontal surface as shown to the right. The coefficient of sliding friction between these two masses and the horizontal surface is m-0.520. A string is attached to the end of mas m,8.00 kg. This string is then looped arounda pulley and is finally attached to the left vertical surface. The pulley is attached to mass m 12.0 kg as shown and then a force F is applied to my sach that m moves toward the right at a constant velocity of 5.0 m/sec. a. What will be the magnitude of the force F needed to pull mass my to the right at a constant speed? b. What will be the tension T in the string? m2 25. Two masses, m-3.0 kg and m; 7.0 kg. are sitting on a horizontal surface as shown to the right. The two masses are attached together by a string in which the tension is T. A force F-67.0 N is applied to the system so as to pull the two masses to the left at a constant speed. a What will be the coefficient of sliding friction for this surface? b. What will be the tension T in the string connecting the two masses together? т m 26. Two masses are arranged as shown. Mass m, has a mass of 4.00 kg and is attached to the vertical surface on the left with a string in which the toension is T. Mass m, 6.00 kg, is sitting on the horizontal surface and is being pulled to the right by a force F so that ma is moving to the right at a constant speed. The coefficient of sliding friction between m, and my is u-0.380 while the coefficient of sliding friction between my and the horizontal surface is u 0.510 a. What will be the tension T in the string? b. What will be the magnitude of the force F required to pull my to the right at a constant speed? m, 27. In the diagram at the right m; 8.00 kg. is sliding down the incline at a constant speed. m, is being pulled by a string attached between these two masses. The coefficient of kinetic friction between m, and the horizontal surface is u,-0.220 while the coefficient of kinetic friction between mi and the incline is u 0.380. The angle between the incline and the horizontal is a 42.0. What is the mass of mi? direction of motion m, 28. The coefficient of kinetic friction between the tires of a car, which has a mass of 50 kg., and the road is -0.630 while the coefficient of static friction is u,- 0.680. How much more frictional force is available to the driver of this car when the car is NOT in a skid? car 29. A scissors jack is a common configuration for a simple jack used to lift a car to replace a flat tire. The diagram at the left shows the basic configuration of such a jack. The force F is generally supplied by a screw which is turned by a lug wrench. Assume that the angle in the diagram is a 15.0 a. What will be the magnitude of the force F required to lift a mass of M-450 kg? b. What will be the magnitude of the force F if the angle is a-85.0? 20d. 0.675 23a. 39.2 N 20c. 143 N 22. 243 lb Answers to opposite side: 20b. 621 N 21b. 79.7 N 23b. 0.315 2ld.0.0679 21c. 1170 N C J. Kovalcin 20

Answer 1

m7 ml M = 12 kg me=0.52 - 5 m/s az o for mass in He 2T F1 fa mmg mg for mass m. mass m to move at constant speed, F = 4+2T = m m g+27= 0.52x 12 x 9.8 & 2T F = 61.152 +2T m. , f - mama For m to move at contant velouty, I t mm.9 -0.52%8x9.8 T = 40.768 N - 61.152 + 2x 40.768 = 142.7 N. Fa 142.7 N "T m = 3kg, m = 7kg fa 67 N for two man to move at constant speed. mu for mann, F = Ttf - Ttm mg = Ttm x 3×9.8 E = T+29.4 m ) for mawm T = of a mam go my x rx 9.8 - 68.6 m x98 68.6 m - (2)

Equating equation (1) and (2), it = 68.6 mx +29.4 m, 98 mx 2 67 m. 67 98 - ma=0.684) 0.684 efficient of sliding friction I= 68.6m, 68.6 x 0.684 2 46.92 N. IT = 46.92 N
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