(a) A bathroom scale has a spring that is compressed by 1.50 cm when a 165 kg person stands on the scale. Assuming the spring is described by Hooke's law, find its force constant.

(b) How much would it be compressed if a 94.0 kg person stands on the scale in an elevator that is accelerating upward at 1.25 m/s2?

(a) Well, if I were a bathroom scale, I'd definitely be springing into action at the sight of a 165 kg person! Let's get down to business and use Hooke's law to find the force constant. Hooke's law states that the force exerted by a spring is directly proportional to its displacement from its equilibrium position.

So, the formula we're looking at is F = -kx, where F is the force, k is the force constant, and x is the displacement. In this case, the displacement is 1.50 cm, which we'll have to convert to meters to keep things consistent.

1.50 cm = 0.015 m

Now, we know that the force exerted by the spring is equal to the weight of the person standing on it. So we have:

F = mg

where m is the mass (165 kg) and g is the acceleration due to gravity (approximately 9.8 m/s²).

F = (165 kg) * (9.8 m/s²)
F = 1617 N

Plugging that back into Hooke's law:

1617 N = -k * 0.015 m

Now we just need to solve for k:

k = -1617 N / 0.015 m
k ≈ -107800 N/m

So, the force constant of the bathroom scale's spring is approximately -107800 N/m.

(b) Now let's move on to the second part of the question, where a 94.0 kg person is standing on the scale in an elevator that is accelerating upward at 1.25 m/s².

In this case, we need to calculate how much the spring would be compressed, considering the additional acceleration. Since the scale itself is in an accelerated frame of reference, we need to account for both the person's weight and the net force acting on them.

We'll start by calculating the net force:

Net force = ma

where m is the mass (94.0 kg) and a is the acceleration (1.25 m/s²).

Net force = (94.0 kg) * (1.25 m/s²)
Net force = 117.5 N

Now, let's find the displacement of the spring using Hooke's law:

F = -kx

117.5 N = -kx

Since we're solving for displacement (x), we can rearrange the equation:

x = -117.5 N / k

Using the force constant we found in the previous part (-107800 N/m), we can calculate the displacement:

x = -117.5 N / (-107800 N/m)
x ≈ 0.0011 m

So, the spring would be compressed by approximately 0.0011 m if a 94.0 kg person stands on the scale in an elevator that is accelerating upward at 1.25 m/s².

(a) To find the force constant of the spring, we can use Hooke's law equation:

F = kx

Where:
F = force applied to the spring
k = force constant of the spring
x = displacement of the spring

Given that a 165 kg person stands on the scale and compresses the spring by 1.50 cm (or 0.015 m), we can rearrange the equation to solve for k:

k = F/x

First, we need to determine the force applied to the spring. The force can be calculated using the person's weight:

F = m * g

Where:
F = force
m = mass of the person
g = acceleration due to gravity (approximately 9.8 m/s^2)

Substituting the given values:

F = 165 kg * 9.8 m/s^2 = 1617 N

Now we can calculate the force constant:

k = 1617 N / 0.015 m ≈ 107800 N/m

Therefore, the force constant of the spring is approximately 107800 N/m.

(b) To find the compression of the spring when a 94.0 kg person stands on the scale in an elevator accelerating upward at 1.25 m/s^2, we can use the same equation as in part (a):

F = kx

First, we need to determine the force applied to the spring. The force can be calculated using the person's weight:

F = m * g

Where:
F = force
m = mass of the person
g = acceleration due to gravity (approximately 9.8 m/s^2)

Substituting the given values:

F = 94.0 kg * 9.8 m/s^2 = 922 N

Next, we need to consider the additional force applied to the spring due to the elevator's upward acceleration. This force can be calculated using Newton's second law:

F = m * a

Where:
F = force
m = mass of the person
a = acceleration

Substituting the given values:

F = 94.0 kg * 1.25 m/s^2 = 117.5 N

Now we can calculate the total force applied to the spring:

Total force = force due to weight + force due to acceleration
Total force = 922 N + 117.5 N = 1039.5 N

Finally, we can calculate the compression of the spring using the force constant:

F = kx

Rearranging the equation to solve for x:

x = F / k

Substituting the given values:

x = 1039.5 N / 107800 N/m ≈ 0.00963 m

Therefore, the spring would be compressed by approximately 0.00963 m if a 94.0 kg person stands on the scale in an elevator that is accelerating upward at 1.25 m/s^2.

(a) To find the force constant of the spring, we can use Hooke's law, which states that the force exerted by a spring is directly proportional to the amount it is compressed or stretched.

Hooke's law is expressed as F = kx, where F is the force exerted by the spring, k is the force constant, and x is the displacement from the equilibrium position.

In this case, we are given the displacement x = 1.50 cm = 0.015 m and the mass m = 165 kg. We need to find the force constant k.

First, we calculate the force exerted by the person using the formula F = mg, where g is the acceleration due to gravity. Since the person is standing still, the weight is the only force acting on them. Thus, F = mg = 165 kg * 9.8 m/s^2 = 1617 N.

Now, we can set up the equation F = kx and solve for k.

1617 N = k * 0.015 m

Dividing both sides of the equation by 0.015 m, we get:

k = 107800 N/m

Therefore, the force constant of the spring is 107800 N/m.

(b) To determine how much the spring will be compressed when a 94.0 kg person stands on the scale in an elevator accelerating upward at 1.25 m/s^2, we need to consider the additional force acting on the person due to the acceleration.

First, we calculate the force exerted by the person using the formula F = mg, where m = 94.0 kg and g = 9.8 m/s^2. Thus, F = 94.0 kg * 9.8 m/s^2 = 921.2 N.

To account for the additional force due to acceleration, we need to calculate the net force acting on the person. The net force is given by the equation F_net = F - ma, where a is the acceleration and m is the mass.

F_net = 921.2 N - (94.0 kg * 1.25 m/s^2) = 921.2 N - 117.5 N = 803.7 N.

Next, we can use Hooke's law F = kx to find the displacement x of the spring.

803.7 N = k * x

Solving for x, we divide both sides of the equation by the force constant k:

x = 803.7 N / 107800 N/m = 0.0075 m

Therefore, the spring would be compressed by 0.0075 m, or 0.75 cm, when a 94.0 kg person stands on the scale in an elevator accelerating upward at 1.25 m/s^2.

(a)

F=k•Δx
F=mg =>
mg=k•Δx
k= mg/Δx =165/0.015 = 11000 N/m
(b)
m₁a=N-m₁g
N=m₁(g+a)
k•Δx₁= m₁(g+a)
Δx₁=m₁(g+a)/k=
=94(9.8+1.25)/11000=0.094 m