There is a vertically suspended spring, mass system. When block of mass 10 kg is suspended from lower end of the spring, it is stretched by 20 cm under the load of block at equilibrium position. When an upward speed of 4 m/s is imparted to the block by giving a sharp impulse from below, how much high will it rise from equilibrium position.
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
(0.56 m = 56 cm)
Sol.
mg = kx
K = 
= 
= 500 N/m
K (0.2) 2 + 
mv 2 = m × 10 × 0.2
× 500 × 4 × 10 –2 + 
× 10 v 2 = 10 × 10 × 0.2
10 + 5v 2 = 20
v 2 = 2
v = 
m/s 
Since u is 4 m/s ( 
) so block will compress the spring.
Let x be the compression of spring.
+ 
+ 0 = 
m (0) 2 + 
Kx 2 + mg (x + 0.2)
× 10 (4) 2 
+ × 500 × 
= 
× 500 (x) 2 + 10 × 10 (x + 0.2)
80 + 10 = 250x 2 + 100 x + 20
25 x 2 + 10 x – 7 = 0 solving this
x = 0.36 m
So, from initial position distance is (0.2 + 0.36) m = 56 cm
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