A 2-kg block is dropped from a height of 0.4 m on a spring of force constant k=1960 N. Find the maximum distance the spring will be compressed. (Take g = 9.8 ms^(-2)) [Hint : Consider conservation of energy]
Answers
Answer:
Mass of the body is ( m ) = 2 Kgspring's constant ( K ) = 1960 N/m height of the body = 0.4 mLet the maximum distance spring will compressed = x m see the attachment , it is clear that initial potential energy of body = mg(h + x) final potential energy = spring potential energy = 1/2 Kx² according to law of conservation of energy,initial potential energy = final potential energy mg(h + x) = 1/2 Kx² 2mgh + 2mgx = Kx²
Kx² - 2mgx - 2mgh = 0
⇒ 1960x² - 2*2*9.8*x - 2*2*9.8*0.4 = 0
⇒ 1960x² - 39.2x - 15.68 = 0
after solving this quadratic equations ,
we get x = 0.1m
hence, answer is x = 0.1m
Answer:
Mass of the body is ( m ) = 2 Kg
spring's constant ( K ) = 1960 N/m
height of the body = 0.4 m
Let the maximum distance spring will compressed = x m
see the attachment , it is clear that
initial potential energy of body = mg(h + x)
final potential energy = spring potential energy = 1/2 Kx²
according to law of conservation of energy,
initial potential energy = final potential energy
mg(h + x) = 1/2 Kx²
2mgh + 2mgx = Kx²
Kx² - 2mgx - 2mgh = 0
⇒ 1960x² - 2*2*9.8*x - 2*2*9.8*0.4 = 0
⇒ 1960x² - 39.2x - 15.68 = 0
after solving this quadratic equations , we get x = 0.1m
hence, answer is x = 0.1m