Question

Two blocks of masses 6 kg and 4kg are connected by a rope of mass 2kg are resting on a frictionless floor as shown in the following figure. If a constant force of 60N is applied to 6kg block, then the tension in the rope is given by point A, B and C are respectively given by

Answer 1

Net force on the system of masses given above is given as

$F = 60 N$

now as per Newton's II law we have

$F = (m_1 + m_2 + m_3)a$

$60 = (6 + 4 + 2)*a$

$a = \frac{60}{12} = 5 m/s^2$

Now tension at the point A is given by Newton's law

$F = ma$

$F_a = (4 +2)* 5 = 30 N$

Similarly we have to find tension at point B

$F_b = (4 + 1)* 5 = 25 N$

Similarly we have to find tension at point C

$F_c = 4* 5 = 20 N$

Now as per above the tension at point A , B , C is 30 N, 25 N, 20 N

Answer 2

Answer:

5m/s. 30n 25n 20n

Explanation:

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