Question

2. The blocks of masses 2 kg, 3 kg and
5 kg are connected by light,
inextensible strings as shown. The
system of blocks is raised vertically
upwards by applying a force
F. = 200 N. Find the common
acceleration and tensions in the
strings.​

Answer 1

Answer:

a-10m/s² T-100

Explanation:

a-100/10 se(200-100)

a-10m/s²

T k liye

dekh equation bana

200=T+100(due to mg)

T=100

Answer 2

Given :-

Three block of masses 2 Kg, 3 Kg and 5 Kg are connected by light, inextensible string respectively.

M = 5 Kg

m = 3 Kg

m' = 2 Kg

Its acceleration will be 10 ms-² i.e. a = 10 ms-².

200 - 100 = (M+m+m')a

100 = 10a

a = 10 ms-².

This question we will solve using two blocks as a system.

For 2 Kg and 3 Kg block taking them as System.

200 - ${T}_{2}$ - (m+m')g = (m+m')a

200 - ${T}_{2}$ - 50 = 50

${T}_{2} = 100 N$

Again,

Taking blocks 3 Kg and 5 Kg as a system.

${T}_{1}$ - (M + m)g = (M+m)a

${T}_{1}$ - 80 = 80

${T}_{1} = 160 N$

Hence,

The tension in first string = ${T}_{1}$ = 160N

The tension in second string = ${T}_{2}$ = 100 N