A spring scale hung from the ceiling stretches by 6.1 cm when a 2.0 kg mass is hung from it. The 2.0 kg mass is removed and repl
aced with a 2.5 kg mass.
Part A What is the stretch of the spring?
1 answer:
To solve this problem, we need to use Hooke's Law.
Hooke's Law:
F = -kx
where F = mg
m = mass of the object
g = gravitational force
k = spring constant
x = length of stretch of the spring
Solve for the spring constant, k:
2.0 kg * 9.81 m/s^2 = -k*(6.1cm/100cm/m)
k = 321.64
The stretch of the spring with mass of 2.5 kg is then determined:
2.5 kg * 9.81 m/s^2 = -321.64 *(x)
x = 7.62 cm
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The Answer is Option C
I think...
Sorry If i am wrong...

Since initial velocity is zero hence , u = 0
=> d = 1/2 * a * t2

on solving we get
d = 86.436 metres
Note ; Here Gravitational Acceleration is take as , g = 9.8 m/s2
You can’t solve it because you don’t have c in the question
Answer:
the answer is true.
Explanation:
hope it will help you