Question

What is the magnitude of the force required to stretch a 20 cm-long spring, with a spring constant of 50 N/m, to a length of 21 cm?

Answer 1

Given :

Length of the spring = 20 cm

Spring constant = 50 N/m

New length = 21 cm

To Find :

Magnitude of applied force which causes elongation in the spring.

Solution :

According to Hooke's law, the force needed to extend or compress a spring by some distance is proportional to that distance.

$\dag\:\underline{\boxed{\bf{\purple{F=k\cdot\Delta x}}}}$

where k denotes spring constant

• Initial length x₁ = 20 cm = 0.2 m

• Final length x₂ = 21 cm = 0.21 m

By substituting the given values;

➠ F = k (x₂ - x₁)

➠ F = 50 × (0.21 - 0.2)

➠ F = 50 × 0.01

➠ F = 0.5 N

Knowledge BoosteR :

• All the central forces are conservative forces.
• For the spring potential energy 1/2 kx², the zero of the potential energy is the equilibrium position of the oscillating mass.
• Every mechanical force is not associated with a potential energy. The work done by friction over a closed path is not zero because no potential energy can be associated with friction.