3. In the Atwood machine the heavier body on left has mass M = 1 kg, while the l

Question

3. In the Atwood machine the heavier body on left has mass M = 1 kg, while the lighter body on right has mass m = 0.9 kg. The system is initially at rest, there is no friction and the mass of the pulley is negligible (a) write down the algebraic expression for the full energy E at the initial instant when the larger mass is at elevation h and the smaller mass is at elevation (b) write down the algebraic expression for the full energy E at the final instant when the larger mass is at elevation 0 and the smaller mass is at elevation h; use the yet unknown speed v when evaluating the full kinetic energy (c) Find the speed after M lowers by h = 50 cm from initial position (d) find the acceleration a (you can either use the above relation between u2 and h, with h = 2/(2a), or use the 2nd Law) (e) find the net force acting on M

Explanation / Answer

a) E = [KE(M) +PE(M)] + [KE(m) + PE(m)]

= [0 + Mgh] + [0 + mg(0)] {KEs are zero since initial velocities are zero}

= Mgh (where g = 9.91m/s2 is the acceleration due to gravity).

b) Now, both the blocks are moving with the same velocity v but in opposite direction. So, the sign of v will be opposite for both.

E = [(0.5*M*v^2 + Mg(0)] + [(0.5*m*(-v)^2) + mgh] {Taking downward v as positive}

= (0.5)(v^2) [ M + m ] + mgh

c) Re-arranging the above eqn to calculate v^2:

v^2 = (E-mgh)/(0.5*(M+m))

By conservation of momentum, initial E = final E

SO, from (a) : E = Mgh = (1x9.81x0.50) = 4.905 J

S0:

v^2 = (4.905 - 0.9*9.81.0.5)/(0.5(1+0.9)) = 0.516

=> v = 0.719 m/s2

d)

Using the given relation:

h = v^2/2a

=> a = v^2/2h = (0.516)/(2*0.50) = 0.516 m/s2

e)

By Newton's 2nd law of motion:

Net force = mass x a

F = Ma = (1)(0.516) = 0.516N

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