A 2kg block is dropped from a heights of 40 cm onto a spring constant k=1960N/m.find the maximum distance the spring bis compressed
Answers
Answer:
Explanation:
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
m = 2 kg, h = 0.4 m, k = 1960 N/m
Let the compression be x. By including it, the potential energy of the block will be,
This provides the potential energy to the spring due to elasticity, i.e.,
We assumed a positive value for x, and hence,