Question

two blocks A and B are connected to each other by string by and the spring string passes over a frictionless Pulley as shown in the figure block B slides the horizontal top surface of earth stationary block c and the block is slide along the vertical side of SIM both with same uniform speed of shaft of friction between the surface of blocks is 0.2 force constant of the spring is 1960 Newtown/m.if the mass of a block Aid 2kg find the mass of the block B find the tension in the spring find the potential energy stored in the spring..
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Answer 1

Answer:

Explanation:

(1) As masses A and B are moving with constant velocity this is a problem of dynamic equilibium i.e., forces acting on mass A (or B) balance each other.

(2)As string and spring are weithless and no mass is involved between them, Tstring=Tspring=T

(3) Force of friction on block B, fB=μRB=μmBg (as RB=mBg) while on block A, fA=μRA=0 (as RA=mAgcos90∘=0)

In the light of above for horizontal equilibrium of B,

T=fB=μmBg ....(i)

While for vertical equilibrium of A,

T=mAg ....(ii)

So, from Eqns (i) and (ii)

μmBg=mAg

mB=(2/0.2)=10kg

Now, as for spring T=ky in the light of Eqn. (ii) becomes

ky=mAgi.e.,y=(2×9.8)/1960=10−2m

So, the enegy stored in the spring

U=(1/2)ky2=(1/2)×1960×(10−2)2=0.098J

Answer 2

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