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3 years ago

Plz help ASAP I'll mark as brainliest

Physics

1 answer:

gogolik [260]3 years ago

8 0

Hi there!

Hooke's law states that:

F = -kx

k = Spring constant (N/m)

x = DISPLACEMENT from equilibrium (m)

Essentially, the force of a spring is PROPORTIONAL to its spring constant and its displacement from its equilibrium point.

The force of the spring (T) is not proportional to the spring's length (l), but rather its DISPLACEMENT from its equilibrium length. (Δl)

The equilibrium length is where the force of the spring (T) = 0N. Looking at the graph, the line intersects this value at l = 30cm.

We can begin by looking at the given graph.

When the spring force = 4N, the total length of the spring is 35 cm.

Now, the EQUILIBRIUM length is 30 cm, so the total elongation is:

35 - 30 = 5 cm.

5.1.

If the spring elongates by 10 cm, the total length of the spring is:

30 + 10 = 40 cm

According to the graph, a length of 40 cm corresponds to a force of 8N.

5.2.

We can solve for the weight of the ball using the following:

W (weight) = m (mass) · acceleration due to gravity (10N/kg)

Using a summation of forces:

∑F = T - W

The elongation that we are solving for occurs at the equilibrium point (net force = 0 N), so:

0 = T - W

T = W = 8 N

5.3.

0 = T - Mg

T = Mg

Use the prior value of T and gravity to solve:

8 = 10M

m = 0.8 kg

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