Block 1, of mass m_1 = 0.700 kg, is connected over an ideal (massless and fricti
ID: 2222290 • Letter: B
Question
Block 1, of mass m_1 = 0.700 kg, is connected over an ideal (massless and frictionless) pulley to block 2, of mass m_2, as shown. For an angle of theta = 30.0^circ and a coefficient of kinetic friction between block 2 and the plane of mu = 0.400, an acceleration of magnitude a = 0.450 m/s^2 is observed for block 2.i found out that Net force on m1 : m1g - T = m1.a ...............(1)
Net force on m2 : T - m2.g.sin ? - ?k.m2.g cos ? = m2.a ............(2) (i don't know how to get net force on m2)
Please solve this problem with a pix that has forces on it so i can understand better! Thank you!
I DONT JUST WANT ANSWER!
Explanation / Answer
You have given the answers without units. [m1– m2 (sin ? + µ cos ?)] g is the pulling force pulling both masses together with an acceleration of a Hence a = [m1– m2 (sin ? + µ cos ?)] g / (m1 + m2) 0.4 = [0.7– m2 (sin 30 + 0.4cos 30)] 9.8 / (0.7+ m2) Solving m2 = 0.757 kg =============================== 0.4*(0.7+ m2) = [6.86 – 8.2908 m2] 0.28 + 0.4 m2 = 6.86 – 8.2908 m2 8.6908 m2 = 6.58 m2= m2 = 0.757 kg