Aforce of 34n stretches a very light ideal spring 0.73 m from equilibrium, what is the

Spring constant, k = 47 N/m

Explanation:

It is given that,

Force applied to a spring, F = 34 N

A very light ideal spring moves 0.73 m from equilibrium position i.e. x = 0.73 m

We have to find the force constant or spring constant of the spring. It can be calculated using Hooke's law. According to him, the force acting on the spring when it compresses or stretches is given by :

[tex]F=-kx[/tex]  (-ve sign shows opposite direction)

[tex]k=\dfrac{F}{x}[/tex]

[tex]k=\dfrac{34\ N}{0.73\ m}[/tex]

k = 46.5 N/m

or

k = 47 N/m

Hence, the spring constant of the spring is 47 N/m.

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gpe = mass x gravity constant x height or… gpe = mgh

where… mass: measured in kilograms (kg)

gravity: gravity constant on earth = 10.00 m/s2

height: measured in meters (m)

gpe: measured in joules (j) or kilojoules (kj)

how do you solve gpe?

gravitational potential energy (gpe) is the energy an object has because of its position above the. ground.

gpe = mass x gravity constant x height or… gpe = mgh.

mass:

note: to earn full marks when solving science word problems, you must show your work:

formula.

example:

given values:

mass = 0.3 kg.