Consider the system shown in the figure. In the initial state, the spring is uns
ID: 1893385 • Letter: C
Question
Consider the system shown in the figure. In the initial state, the spring is unstretched and is attached to the m_1 = 8.00 kg block on one end and a wall on the other. In turn the 8.00 kg block is attached via a light rope passing over a massless pulley (with frictionless axle) to the m_2 = 21.5 kg block situated on a heta = 31.0^circ incline. The coefficient of kinetic friction between both masses and the surfaces is mu_k = 0.250, and the spring constant is k = 90.0 N/m.
Explanation / Answer
You have 0% respect score. I hope you will rate my answers Part A When the spring is stretched the force is acting towards right. From the free body diagram we assume that acceleration of masses is 'a'. We get the following equations T - 21.5g*sin(31) - 21.5 *g*cos(21)*.25 = 21.5 *a 90*.3 - 8*g*.25 - T = 8a adding the two equations 27 -19.6 -108.51- 45.1512875 = a(29.5) a = -146.2/29.5 = -4.958009 The accelration is coming negative which should not have happned as per our convention of direction. So it means that the block must be heavy enough for the spring to pull. We can check 21.5g*sin(31) >90*.3 108.51 >27 so the blocks do not move at all. so , a = 0 B) Tension can be found out using the above equations 90*.3 - 8*g*.25 - T = 8a T = 90*.3 - 8*g*.25 = 7.4 N C) the system is always in translational equlibrium so the extension is the same as 0.3 m