A 10-kg object,hung by an ideal spring,causes the spring to stretch 2.0 cm.What is the spring constant for this spring?
A)5.0 N/cm
B)49 N/cm
C)0.20 N/cm
D)20 N/m
E)0.0020 N/cm

Question

Quizplus · Accepted Answer

We can use Hooke's Law which states that the force exerted by a spring is proportional to its displacement from equilibrium. Mathematically, this can be written as F = -kx, where F is the force, x is the displacement, and k is the spring constant.

In this case, we know the mass of the object (10 kg) and the displacement of the spring (2.0 cm = 0.02 m). The force exerted by the spring can be calculated using the formula F = mg, where g is the acceleration due to gravity (9.81 m/s^2).

F = 10 kg * 9.81 m/s^2 = 98.1 N

Now we can use Hooke's Law to find the spring constant:

98.1 N = -k * 0.02 m

k = -98.1 N / 0.02 m

k = -4905 N/m

However, since the displacement x is in centimeters instead of meters, we need to convert the units:

k = -4905 N/m * (1 m / 100 cm)^2

k = -49.05 N/cm

But since the question is asking for a positive value of the spring constant, we can take the absolute value:

k = 49 N/cm

Therefore, the answer is B.

A 10-Kg Object,hung by an Ideal Spring,causes the Spring to Stretch