If a spring has a constant of #8 (kg)/s^2#, how much work will it take to extend the spring by #40cm #?

Question

If a spring has a constant of #8 (kg)/s^2#, how much work will it take to extend the spring by #40cm     #?

Sebastian Acosta · Accepted Answer

Consider your system,

Recall Hooke's Law,

#F_"s" = -kx#, and by extension

#F_"ext" = -F_"s" = kx# (we're looking to find this)

where #k# is the constant you describe and #x# is the displacement of spring.

Moreover, recall that work is represented by,

#W = F*d#

Now,

#=> W = (1/2kx)*x = 1/2kx^2#

(the reason this is halved is because we're averaging the force)

Hence,

#W_"ext" = 1/2 * (8"kg")/("s"^2) * (0.40"m")^2 approx 0.64"J"#

To be sure, the elastic potential energy is equal to the work done to stretch it that far (which shares the same equation as that derived).

Benjamin Asiedu · Answer

undefined To calculate the work required to extend the spring by 40 cm, you can use the formula for the work done on a spring:

$$ W = \frac{1}{2} k x^2 $$

Where:
- $ W $ is the work done,
- $ k $ is the spring constant,
- $ x $ is the displacement from the equilibrium position.

Given:
- $ k = 8 \, \text{kg/s}^2 $,
- $ x = 40 \, \text{cm} = 0.4 \, \text{m} $,

Substitute the values into the formula:

$$ W = \frac{1}{2} \times 8 \times (0.4)^2 $$

$$ W = 0.5 \times 8 \times 0.16 $$

$$ W = 0.5 \times 1.28 $$

$$ W = 0.64 \, \text{J} $$

So, it will take 0.64 joules of work to extend the spring by 40 cm.