A block of mass 1.70 kg is accelerated across a rough surface by a light cord pa

Question

A block of mass 1.70 kg is accelerated across a rough surface by a light cord passing over a small pulley as shown in the figure below. The tension T in the cord is maintained at 10.0 N, and the pulley is 0.120 m above the top of the block. The coefficient of kinetic friction is 0.340.
(a) Determine the acceleration of the block when x = .400 m.

(c) Find the maximum value of the acceleration and the position x for which it occurs.

(d) Find the largest value of x for which the acceleration is zero.

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Explanation / Answer

T= 10 N
force in horizontal direction = Tcos y = 10 X 0.12/sqrt[0.12^2 + x^2 ]
force in vertical direction = Tsin y = 10X x/sqrt[0.12^2 + x^2 ]
now, accn

a = F - Ff /m

F= Tcosy
Ff= frictional force =N = ( mg - Tsiny ) .

a.now substute x= 0.4 and fined the accn.

b.differentiate da/dx

. find the maximum value by equating it to zero(da/dx=0) and then put the x back in original eqn.

c. put a= 0 and now compute the value of x.

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