2KG BLK IS ATTACHTED TO HORIZONTAL SPRING WITH A SPRING OMSTANT 200N/M THE BLOCK IS GIVEN A SPEED OG5M/S THEN ELONGATION OF SPRING

Question

ITS FRM WORK ENERGY AND POWER

Answer 1

m = 2 kg

v = 5 m/s

k = 200 N/m

KE_i = 0J
PE_i = (1/2)kx² ←elastic PE
KE_f = (1/2)mv²
PE_f = 0J

 

KE_i + PE_i = KE_f + PE_f

⇒ 0J + (1/2)kx² = (1/2)mv² + 0J

⇒ x² = (mv²)/k

⇒ x = ((mv²)/k)^0.5 = (((2 kg)(5 m/s)²)/(200 N/m))^0.5

= 0.5 m