Question

Two bloks of masses 20 kg and 50 kg are lying on a horizontal floor (coefficient of friction = 0.5).Initially string is stretched andblocksare at rest : Now two forces 300N and 150N is applied on two blocks as shown in figure. The aeleration of 20kg block is 2.5K m/s2 . Find the value of K.

Answer 1

SOLUTION:-
Given by:-
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》Let the tension in the rope = T
》accleration of masses =a
》The frictional force acting on 50kg block is
》And that on 20 kg block is 100N.
Therefore,
》we can say that,
》300 - (250 + T) = 50a
》300- 250 +T = 50a
》50-T = 50a ...........(1)
and,
》150 + T - 100 = 20a
》50+T = 20a............(2)
》add eq(1) and (2)
》 we get,
》100= 70a
》 a = 100/70
here , 2.5 k = 100/70
$2.5k = \frac{100}{70} = \frac{10}{7} \\ k = \frac{10}{7 \times 2.5} \\ k = \frac{4}{7}$

[HENCE K = 4/7 ]

■I HOPE ITS HELP■

Answer 2

For the block of 50 kg....
Friction=0.5×50×10=250N...
So, Tension in rope=50 N...
BT the applied force on 20kg block is 150N and 50 N tension is also applied in same direction.. Therefore, 150 N will act...
Frictional force of 20 kg block=0.5×20×10= 100 N
Therefore. Acceleration= resultant force/mass= 150-100/20=2.5m/s^2.....