A person who weighs 715 N steps onto a spring scale in the bathroom. And the spring compress by 0.830 cm. (a) what is the spring constant? (b) what is the weight of another person who compress the spring by 0.365 cm?
To find the spring constant, we can use Hooke's Law:
F = -kx
Where:
F is the force applied to the spring
k is the spring constant
x is the displacement of the spring.
(a) To find the spring constant:
Given:
Force (F) = 715 N
Displacement (x) = 0.830 cm = 0.0083 m (converting to meters)
k = ?
Using Hooke's Law:
715 N = -k * 0.0083 m
Rearranging the equation to solve for k:
k = -715 N / 0.0083 m
k ≈ -86145.783 m^(-1)
Therefore, the spring constant is approximately -86145.783 m^(-1).
(b) To find the weight of another person:
Given:
Displacement (x) = 0.365 cm = 0.00365 m (converting to meters)
k = -86145.783 m^(-1)
Weight = ?
Using Hooke's Law:
Weight = -k * x
Weight = -(-86145.783 m^(-1)) * 0.00365 m
Weight ≈ 314.677 N
Therefore, the weight of another person who compresses the spring by 0.365 cm is approximately 314.677 N.