Question

System shown in figure is released from rest. Pulley and spring is
massless and friction is absent everywhere. The speed of 5 kg block
when 2 kg block leaves contact with ground is : (Take force con-
stant of spring k = 40 N/m and g = 10 m/s)
(A) √2 m/s
(B) 2√2 m/s
C) 2 m/s
(D) 4√2 m/s​

Answer 1

The speed of the block is v= 2 √2 m/s

Explanation:

Let x be the extension in spring when 2 kg mass leaves contact with the ground.

kx=2g

x = 2g / k

=2×10 / 40= 1 / 2 m

By law of conservation of energy for 5 kg mass

mgx = 1 / 2k^2 + 1 /2 mv^2

v = √2gx − kx^2 /m

v =√2 × 10 × 1 /2 − 40 / 5 × 14

v= 2 √2 m/s

Hence the speed of the block is v= 2 √2 m/s

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Answer 2

Answer:

hey...pls stop copying

juz kidding..