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University Physics Vo D Fisica para Ciencias e × D Physics for Scientists a LON CAPA Falling ana Q LON CAPA Siding MeRX e The Position Of A Mesx x X x a -> C Dloncapa.hep.uprm.edu/enc/72/f187e4e90472cc0ac38372eea272b3e9a2c1d8f5e8b537354bc90abcd72d3b9039b2d28d3b8ba1f57b10e2a49a7523e28cf2c5cedd9b519ff1 84619e4da7f8f8fe ERICK I PEREZ(Student section: 011) Main Menu Contents Grades course contents» » Cap5 Leyes de Newton Fisi3171-Agosto-2017 Messages Courses Help Logout Falling and sliding Mass >, Timer Note., Evaluate 4-Feedback-Print Falling and Sliding Mass Points:1 Force Mass m1=25.9 kg is on a horizontal surface, connected to mass m2= 6.90 kg by a light string as shown. The pulley has negligible mass and no friction. A force of 204.1 N acts on m1 at an angle of 33.9 2 The coefficient of kinetic friction between mi and the surface is 0.213. Determine the upward acceleration of m2 There are many forces involved. The tenson, (on mi and maj. the frictional force on mi, and for which you must first find the nomal force), the horsontal component of F (on mi) and oravity (on m2). Newton's 2nd Law applies. Note both masses have the same size acceleration involved. The tension, (on mi and m2), the frictional force (on m1, and for which you must first find the normal force), the horixontal component of F (on m) and gravity Tries 7/40 Previous Tries Post Discussion Send Feedback Activate Windows Go to Settings to activate Windows 11:49 AM O Type here to search 11/20/2017

Explanation / Answer

Here ,

let the normal force at m1 is N

N = 25.9 * 9.8 - 204.1 * sin(33.9 degree)

N = 140 N

f = 0.213 * 140 = 29.8 N

let the acceleration of the block m1 is a

(m1 + m2) * a = F * cos(theta) - 6.90 * 9.8 - f

(25.9 + 6.90) * a = 204.1 * cos(33.9 degree ) - 6.90 * 9.8 - 29.8

solving for a

a = 2.19 m/s^2

the upwards acceleration is 2.19 m/s^2

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