Question

Q2. Two objects of unequal masses, M-5kg and m 2 kg are connected by a very light cord passing over an ideal pulley of negligible mass. When released, the system accelerates with acceleration a, and friction is negligible. The tension of the string is T Draw one Free Body Diagram for each object. Draw them on above diagram 5 points a. b. For each object and each direction, write down F-m5 points c. Solve the set of equations and find the acceleration of the boxes. 5 points d. Find tension of the code? 5 points

Answer 1

aml m am Mg Mg mg mg (b) For mass M. F+Mg-T = 0 EF=T - Mg Ma=T - Mg For mass m. EF-mg +T = 0 EF = mg-1 ma= mg-1

(c) From part b the acceleration of the box having mass M is, a=- Ma=(mg-ma) - Mg (M +m)a=(m_M)g (m-M)g (M+m) (2 kg - 5 kg)(10 m/s) 2 kg+5 kg = -4.28m/s2 (d) The tension of the code is, Ma=T - Mg T = M(a+g) = (5 kg)(-4.28 m/s +10 m/s) = 28.6 N