A certain spring stretches 5.3 cm when it

supports a mass of 0.65 kg .
If the elastic limit is not reached, how far
will it stretch when it supports a mass of
10 kg ?Answer in units of cm

To find out how far the spring will stretch when supporting a mass of 10 kg, we can use Hooke's Law, which states that the force applied to a spring is directly proportional to the extension or compression of the spring.

Hooke's Law can be 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

To solve this problem, we need to find the spring constant (k) first. We can do this by using the given information that the spring stretches 5.3 cm when supporting a mass of 0.65 kg.

We know that the weight (force) is equal to the mass multiplied by the acceleration due to gravity (9.8 m/s^2).

Weight = mass × acceleration due to gravity
F = m × g

F = 0.65 kg × 9.8 m/s^2
F ≈ 6.37 N

Using Hooke's Law, we can calculate the spring constant (k).

k = F/x
k ≈ 6.37 N / 0.053 m
k ≈ 120 N/m

Now that we have the spring constant, we can find out how far the spring will stretch when supporting a mass of 10 kg.

F = m × g
F = 10 kg × 9.8 m/s^2
F = 98 N

Using Hooke's Law again:

F = kx
98 N = 120 N/m × x

Rearranging the equation to solve for x:

x = F / k
x = 98 N / 120 N/m
x ≈ 0.817 m

Converting the answer to centimeters (cm):

x ≈ 0.817 m × 100 cm/m
x ≈ 81.7 cm

Therefore, when the spring supports a mass of 10 kg, it will stretch approximately 81.7 cm.