A spring that has a spring constant of 150 N/m is oriented vertically with one end on the ground.
(a) What distance does the spring compress when a 4-kg object is placed on its upper end? in m.
(b) By how much does the potential energy of the spring increase during the compression? in J.
To find the distance the spring compresses when a 4-kg object is placed on its upper end, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position. Hooke's Law can be expressed as follows:
F = -kx
where F is the force exerted by the spring, k is the spring constant, and x is the displacement of the spring.
(a) To find the distance the spring compresses, we need to determine the force exerted by the spring when the 4-kg object is placed on its upper end. The force exerted by the object is given by the equation:
F = mg
where m is the mass of the object and g is the acceleration due to gravity.
With m = 4 kg and g = 9.8 m/s², we get:
F = (4 kg) * (9.8 m/s²) = 39.2 N
Now we can rearrange Hooke's Law to find the displacement x:
x = -F / k
Substituting the values, we get:
x = -(39.2 N) / (150 N/m) = -0.261333... m
The negative sign indicates that the spring is compressed.
Therefore, the spring compresses by approximately 0.261 meters when a 4-kg object is placed on its upper end.
(b) To find the increase in potential energy of the spring, we can use the formula for potential energy stored in a spring:
PE = (1/2)kx²
where PE is the potential energy, k is the spring constant, and x is the displacement of the spring.
Substituting the values, we get:
PE = (1/2) * (150 N/m) * (-0.261 m)² ≈ 5.096 J
Therefore, the potential energy of the spring increases by approximately 5.096 joules during the compression.
a. Wo = M*g = 4 * 9.8 = 39.2 N.
d = 39.2N. * 1m/150N. = 0.261 m.