Question

An elevator car of mass 805 kg falls from rest 4.00 m, hits a buffer spring, and then travels an additional 0.500 m, as it compresses the spring by a maximum of 0.500 m. What is the force constant of the spring

Answer 1

To find the force constant of the spring, we can use Hooke's Law. In this case, the elevator car falls 4.00 m and compresses the spring by 0.500 m. The force constant of the spring is 4,808 N/m.

Step-by-step explanation:

To find the force constant of the spring, we can use Hooke's Law, which states that the force exerted by a spring is proportional to its displacement. In this case, the elevator car falls 4.00 m and compresses the spring by 0.500 m. At the point of maximum compression, the spring exerts a restoring force equal to the weight of the car. We can set up the equation F = kx, where F is the force exerted by the spring, k is the force constant of the spring, and x is the displacement of the spring. We can substitute the weight of the car for F and the displacement of the spring for x to solve for k.

First, calculate the weight of the car using the formula w = mg, where m is the mass of the car and g is the acceleration due to gravity. Then, substitute the weight and displacement values into the equation to solve for k.

The force constant of the spring is 4,808 N/m.

Answer 2

2665.6 N/m

Step-by-step explanation:

The potential energy of elevator will convert to PE of the spring

Here,

m=8.5 kg

h=4.0 m

g=9.8m/s2

PE of elevator =mgh = 8.5×9.8×4.0=333.2 J

PE of spring = 1/2 k
`x^(2)`

where,

k=? (to be found)

x= 0.5 m (distance by compressing the spring)

==> PE of spring = 1/2× k× 0.5×0.5 = 0.125k

Since .

PE of elevator = PE of the spring

==> 333.2 =0.125k

==> k = 333.2÷0.125=2665.6 N/m