Two blocks are connected by a light string that passes over a f distance h above

Question

Two blocks are connected by a light string that passes over a f distance h above the floor. pulley as in the figure below. The systern is from rest while m2 is on the floor and m is a (a) Assuming m> m2, find an expression for the speed of ms just as it reaches the floor. (Use any variable or symbol stated above along with the following as necessary: g) (b) Taking mi. 6.3 kg. m2-4.4 kg, and h 4.0 m, evaluate your answer to part (a). m/s (c) Find the speed of each block when m has fallen a distance of 3 3 m m/s Need Help? Read I

Explanation / Answer

a) Apply conservation of mechanivcal energy

gain in kinetic energy = loss potential energy

(1/2)*(m1 + m2)*vf^2 = (m1 - m2)*g*h

vf = sqrt(2*(m1 - m2)*g*h/(m1+m2) )

b) vf = sqrt(2*(6.3 - 4.4)9.8*4/(6.3 + 4.4) )

= 3.73 m/s

c) here we take, h = 1.3 m

vf = sqrt(2*(6.3 - 4.4)9.8*1.3/(6.3 + 4.4) )

= 2.13 m/s

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