Question

Answer with steps .
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class 11th ( NLM ) ​

Answer 1

5a + T − 5g = 0

From free-body diagram (1),

T = 5g − 5a             …..(i)

1/2T-4g-8a=0

⇒ T − 8g − 16a = 0

(1)

From free-body diagram (2),

T = 8g + 16a                    ……(ii)

From equations (i) and (ii), we get:

5g − 5a = 8g + 16a

⇒21a=-3g-a=-9/7

So, the acceleration of the 5 kg mass is  97  m/s2 upward

and that of the 4 kg mass is

2a=2g/7 downward.

(b) From free body diagram-3,

4a-T/2=0

⇒ 8a − T = 0

⇒ T = 8a

Again, T + 5a − 5g = 0

From free body diagram-4,

8a + 5a − 5g = s0

⇒ 13a − 5g = 0

⇒a=5g/13           downward

Acceleration of mass 2 kg is

2a=10/13 (g)and 5 kg is  5g/13.

(c) T + 1a − 1g = 0

From free body diagram-5

T = 1g − 1a                        …..(i)

Again, from free body diagram-6,

T/2-2g-4a=0

⇒ T − 4g − 8a = 0                   …..(ii)

From equation (i)

1g − 1a − 4g − 8a = 0

⇒a=g/3   downward

∴Acceleration of mass 1 kg is  g/3 upward,

Acceleration of mass 2 kg is 2g/3  downward

Answer 2

Answer:

a = 2m/s² ( given ) .

N 1 = 24 N

N 2 = 60 N

N 3 = 120 N .