A spring (k =819N/m) is hanging from the ceiling of an elevator, and a 5.1-kg object is attached to the lower end. By how much does the spring stretch (relative to its unstrained length) when the elevator is accelerating upward at a = 0.46m/s^2?

Torque? How do you think that?

forceonspring=kx=m(a+g)
x=1/k (m(acceleration +g)
solve for x

To find the amount by which the spring stretches, we need to use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position.

Hooke's Law can be stated as: F = -kx

Where:
F is the force applied by the spring (in N)
k is the spring constant (in N/m)
x is the displacement from the equilibrium position (in m)

First, let's calculate the weight of the 5.1-kg object. The weight is the force due to gravity acting on the object and can be calculated using the formula:

Weight = mass * acceleration due to gravity

Given:
Mass of the object (m) = 5.1 kg
Acceleration due to gravity (g) ≈ 9.8 m/s^2

Weight = 5.1 kg * 9.8 m/s^2
Weight ≈ 49.98 N

Since the elevator is accelerating upward, we need to consider the additional force exerted on the object due to this acceleration. The net force can be calculated as:

Net force = F - Weight = ma

Given:
Acceleration of the elevator (a) = 0.46 m/s^2

Net force = ma
Net force = (5.1 kg) * (0.46 m/s^2)
Net force ≈ 2.346 N

Now, we can use Hooke's Law to find the displacement of the spring. Rearranging the formula, we have:

F = -kx
x = -F / k

Substituting the values we have:

x = -2.346 N / 819 N/m
x ≈ -0.00286 m

Therefore, the spring stretches by approximately 0.00286 meters (or 2.86 mm) relative to its unstrained length when the elevator is accelerating upward at 0.46 m/s^2.