A block of wood is compressed 2.0 nm when inward forces of magnitude 120N are applied to it on two opposite sides. (a) Assuming Hooke's law holds, what is the effective spring constant of the block? (b) Assuming Hooke's las still holds, how much is the same block compressed by inwrard forces of magnitude 480N? (c) How much work is done by the applied forces during the compression of part (b)?

k = 120/2*10^-6 N/m

b) four time as far

W = (1/2) k x^2 = (1/2)(120/2*10^-6)(8*10^-6)^2

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

F = k * x

Where F is the applied force, k is the spring constant, and x is the displacement.

Given:
Applied force (F) = 120N
Displacement (x) = 2.0 nm

Plugging the values into the equation, we can solve for the spring constant (k):

120N = k * 2.0 nm

To convert nanometers (nm) to meters (m), we divide by 1 billion (1,000,000,000):

120N = k * 2.0 × 10^(-9) m

Rearranging the equation to solve for k, we get:

k = 120N / (2.0 × 10^(-9) m)
k = 6.0 × 10^10 N/m

Therefore, the effective spring constant of the block is 6.0 × 10^10 N/m.

b) Using the same equation, we can calculate the new displacement (x) for the block when an inward force of 480N is applied:

F = k * x

480N = (6.0 × 10^10 N/m) * x

Solving for x:

x = 480N / (6.0 × 10^10 N/m)
x = 8.0 × 10^(-9) m

The block is compressed by 8.0 × 10^(-9) meters when an inward force of magnitude 480N is applied.

c) The work done by the applied forces can be calculated using the formula:

Work (W) = (1/2) * k * x^2

Given:
Applied force (F) = 480N
Spring constant (k) = 6.0 × 10^10 N/m
Displacement (x) = 8.0 × 10^(-9) m

Plugging in the values, we can calculate the work done:

W = (1/2) * (6.0 × 10^10 N/m) * (8.0 × 10^(-9) m)^2

Calculating the value:

W = 2.4 × 10^(-6) J (in joules)

Therefore, the applied forces do 2.4 × 10^(-6) joules of work during the compression in part (b).

(a) To determine the effective spring constant of the block, we can use Hooke's Law, which states that the force required to compress or extend a spring is proportional to the displacement from its equilibrium position.

Hooke's Law is expressed as:
F = -kx

Where:
F is the force applied to the spring,
k is the spring constant,
x is the displacement of the spring from its equilibrium position.

In this case, the block of wood is behaving like a spring when compressed. We have the following information:

Force applied (F) = 120 N
Displacement (x) = 2.0 nm = 2.0 × 10^(-9) m

Plugging these values into Hooke's Law equation, we get:
120 N = -k * (2.0 × 10^(-9) m)

We can solve for the spring constant (k):
k = -120 N / (2.0 × 10^(-9) m)

Thus, the effective spring constant of the block is -6.0 × 10^10 N/m.

Note: The negative sign indicates that the force applied and the displacement are in opposite directions.

(b) Now, let's find out how much the same block is compressed with an inward force of 480 N using Hooke's Law. We already know the spring constant (k) from part (a).

Force applied (F) = 480 N

We can rearrange Hooke's Law equation to solve for x:
F = -kx
x = -F / k

Substituting the values:
x = -480 N / (-6.0 × 10^10 N/m)
x = 8.0 × 10^(-9) m = 8.0 nm

Therefore, the same block would be compressed by 8.0 nm with an inward force of 480 N.

(c) To calculate the work done by the applied forces during the compression in part (b), we can use the formula for work:

Work (W) = Force (F) * Distance (d)

In this case:
Force (F) = 480 N
Distance (d) = displacement (x) = 8.0 × 10^(-9) m

Therefore:
W = 480 N * 8.0 × 10^(-9) m

Evaluating this,
W = 3.84 × 10^(-6) J

Thus, the work done by the applied forces during the compression in part (b) is 3.84 × 10^(-6) Joules.