Question

A spring force constant 150Nm-1(raised to the power of -1) is acted upon by a constant force of 85N. Calculate the potential energy stored in the spring. (a) 24.08 J (b) 28,00J (c) 0.675J (d) 26.50J

Answer 1

GiveN :

• Force Constant (k) = 150 N/m
• Force Applied (F) = 85 N

To FinD :

• Potential Energy stored in spring.

SolutioN :

Use formula for force :

⇒F = -kx

⇒85 = -150 * x

⇒x = 80/-150

⇒x = - 0.5667

⇒x ≈ - 0.567

But Displacement can never be negative, So, we will take it as positive only.

⇒x ≈ 0.567

∴ Displacement is 0.567 m. (approx.)

______________________

Now, use formula for Potential energy of springs :

⇒P.E = ½kx²

⇒P.E = ½ * 150 * (0.567)²

⇒P.E = 75 * 0.321489

⇒P.E ≈ 24.1

∴ Potential Energy stored is 24.01 J

So, answer is (a) 21.08. Because 21.08 is approximately equal to 21.1 J

Answer 2

Given that ,

Spring constant (k) = 150 N/m

Force on a spring (F) = 85 N

We know that , the force on a spring is given by

$</p><p> \large \sf \fbox{F = - kx</p><p>}$

Where ,

k = spring constant

x = compression

Note : The negative sign indicates the force is opposite to the compression

Thus ,

$\sf \mapsto 85 = - 150 \times x \\ \\\sf \mapsto x = -\frac{85}{150} \\ \\ \sf \mapsto x = -0.5 m$

=> igone the negative value of x

Now , the potential energy of a spring is given by

$\large \sf \fbox{Potential \: energy = \frac{1}{2} k {x}^{2} }$

Thus ,

$\sf \mapsto Potential \: energy = \frac{1}{2} \times 150 \times {(0.5)}^{2} \\ \\ \sf \mapsto Potential \: energy = 75 \times \frac{25}{100} \\ \\ \sf \mapsto Potential \: energy = 18.75 \: \: joule$

$\sf \therefore \underline{The \: potential \: energy \: of \: a \: spring \: is \: 18.75 \: J}$