A spring has an equilibrium length of 20.0 cm and a spring constant of 57.9 N/m. The spring is connected to the underside of the roof of a car and a 0.294 kg block suspended from it. How long (in cm) is the spring when the car is at rest?

How long is the spring if the car is accelerating horizontally at 1.63 m/s2?

L=0.2 m

(a) mg=kx₁
L₁=L+ x₁=L+ (mg/k)=…
(b) m•sqrt(g²+a²)=kx₂
L₂=L+x₂= L + m•sqrt(g²+a²)/k = …

To find the length of the spring when the car is at rest, we can use Hooke's law. Hooke's law states that the force exerted by a spring is directly proportional to its displacement from its equilibrium position. The formula for Hooke's law is:

F = -kx

Where F is the force exerted by the spring, k is the spring constant, and x is the displacement from the equilibrium position.

When the car is at rest, the block is in equilibrium, which means the force exerted by the spring is equal to the weight of the block.

The weight of the block can be calculated using the formula:

F = mg

Where F is the force due to gravity, m is the mass of the block, and g is the acceleration due to gravity (which is approximately 9.8 m/s^2).

Let's calculate the force exerted by the spring:

F = mg = (0.294 kg)(9.8 m/s^2) = 2.8812 N

Now we can use Hooke's law to find the displacement of the spring:

F = -kx

2.8812 N = (-57.9 N/m) x

To solve for x, we rearrange the equation:

x = (-2.8812 N)/(-57.9 N/m) = 0.0498 m

To convert the displacement to centimeters, we multiply by 100:

x = 0.0498 m * 100 = 4.98 cm

Therefore, the length of the spring when the car is at rest is 4.98 cm.

Now let's calculate the length of the spring when the car is accelerating horizontally at 1.63 m/s^2.

When the car is accelerating, there is an additional force acting on the block due to the acceleration of the car. This force can be calculated using Newton's second law:

F = ma

Where F is the net force on the block, m is the mass of the block, and a is the acceleration of the car.

The net force on the block is the sum of the force exerted by the spring and the force due to the acceleration of the car:

Fnet = Fspring + Facceleration

To find the displacement of the spring in this case, we need to consider the net force on the block.

Since the spring is connected to the underside of the roof of the car, the net force on the block is the force exerted by the spring minus the force due to the acceleration of the car (opposite in direction to the spring force).

Let's calculate the force exerted by the spring:

Fspring = -kx

Fspring = -(57.9 N/m)(0.0498 m) = -2.8812 N

Now let's calculate the force due to the acceleration of the car:

Facceleration = ma = (0.294 kg)(1.63 m/s^2) = 0.4782 N

The net force is then:

Fnet = Fspring + Facceleration = -2.8812 N + 0.4782 N = -2.403 N

Now we can use Hooke's law to find the displacement of the spring:

Fnet = -kx

-2.403 N = (-57.9 N/m) x

To solve for x, we rearrange the equation:

x = (-2.403 N)/(-57.9 N/m) = 0.0416 m

To convert the displacement to centimeters, we multiply by 100:

x = 0.0416 m * 100 = 4.16 cm

Therefore, the length of the spring when the car is accelerating horizontally at 1.63 m/s^2 is 4.16 cm.