Ans a) 12(5gd) b)P227m2g3 A block B of mast M is held at rest .pdf

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1) A block B is held at rest on a horizontal table by friction. A light string passes over a pulley and is attached to a hanging mass m which is half the mass of B and is in equilibrium. 2) When released, a constant force P is applied directly to block B as it moves a distance x on the table. The speed of B is v. 3) It is shown that until the string becomes slack, the rate of change of speed (dv/dx) of B equals P/mv^2. Also, when the string becomes slack, the speed v of B equals P/3mg. 4) The distance x that B has moved when the string becomes

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Ans a): 1/2*(5gd) b)P2/27m2g3 A block B of mast M is held at rest on a rough horizontal table.
The coefficient of friction between B and the table is 1/2. One end of a light inextensible string is
attached to B. The string passes over a small smooth pulley fixed at the edge of the table and is
attached to a panicle Q at mass m which hangs freely in equilibrium, as shown in the diagram. It
is given that m = 1/2 M. The system it released from rest. Find the speed of the block when it has
moved a distance d on the table. It is given instead that m = 1/2 M. The system it released from
rest and immediately a force, acting directly towards the policy, is applied to the block. This
force acts with constant power P When B has moved a distance x. its speed is v. The string
becomes slack before B reaches the pulley. Show that until the string becomes slack, dv/dx =
P/mv2" at the instant when the string becomes slack, v = P/3mg" Find the distance that B has
moved at the instant when the string becomes slack.
Solution
a)9.81(3M) - T = (3M)a
29.43M - T = 3Ma
T=29.43M - 3Ma----eq1
T - FR = Ma
FR=RN
=(1/2)(9.81M) N
T - (4.905M) = Ma
T=Ma + 4.905M----eq2
Solve both eqs simultaneously for acceleration a m/s2
29.43M - 3Ma=Ma + 4.905M
29.43M - 4.905M=4Ma
24.525M = 4Ma
a=6.13 m/s2
2ad = v2 - u2
2ad = v2 - 0
v2=2(6.13)d
v2=12.26d
v=3.5d m/s
v=1/25gd m/s

Ans a) 12(5gd) b)P227m2g3 A block B of mast M is held at rest .pdf

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