A block of mass m 1 = 6.31 kg on a frictionless plane inclined at angle = 25.4°

Question

A block of mass m1 = 6.31 kg on a frictionless plane inclined at angle = 25.4° is connected by a cord over a massless, frictionless pulley to a second block of mass m2 = 2.18 kg hanging vertically (Fig. 5-52). (a) What is the acceleration of the hanging block (choose the positive direction up)? (b) What is the tension in the cord?

Fig. 5-52
Problem 57.

Explanation / Answer

Here is my solution: Given : A. m1 = 6.31 kg , m1g (weight) = 61.838 N B. m2 = 2.18 kg, m2g (weight) =21.364 N C. Theta = 25.4 ° D. g = 9.8 m/s^2 Relevant Equations: Fnet = m x a mgsin0 = force of gravitational down a plane 1. Treat each block as two different free body. The higher mass means the direction of acceleration is positive. By drawing the free-body diagrams we should come up with these equations listed below. Block down the plane. 1. m1gsin0 - T = m1 x a 2. T = [ (61.838 N) x sin 25.4 ° ] - [ 6.31a ] 3 .T = (26.5) - 6.31a Block hanging. 1. T-m2g = m2 x a (Plug the equation found above in this equation we get): 2.(26.5) - 6.31a - (21.364) = 2.18a 3. -> Solve for a = .604 m/ s ^2 FINALLY Plug the acceleration found in any equation above to find T. Ex. 1. T = (26.5) - 6.31a 2. T = (26.5) - 6.31(.604 m/s ^2) 3. T = 100.99786 N

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