A block of mass 2kg is dropped from the height of 40 cm on a spring whose force constant is 1960N/m what will be the maximum distance x through which the spring is compressed???
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Answered by
49
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
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
Answered by
10
Answer:
Mass of block m = 2 kg,
h= 40/100 m = 0.4 m,
g = 9.8 ms^−2
k= 1960 N/m.
Work done by the block in compressing the spring = Change in potential energy of the block.
1/2kx^2 = mg(h +x)
[ x = compression produced in the spring]
or,
1/2×1960× x^2 = 2 x 9.8 (0.4 +x)
or, 50 x^2= 0.4 + x
or, x = 10/100 m
= 0.1 m
=10 cm
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