A force of 350 newtons stretches a spring 30 centimeters. How much work is done in stretching the spring from 20 centimeters to 50 centimeters

52.5 J

Explanation:

Applying,

Hook's law,

F = ke Equation 1

Where F = Force, k = spring constant, e = extension.

make k the subject of the equation

k = F/e Equation 2

From the question,

Given: F = 350 Newtons, e = 30 cm = 0.3 m

Substitute these values into equation 2

k = 350/0.3 N/m

Also,

W = 1/2(ke²) Equation 3

Where W = work done in stretching the spring.

Also given: e = (50-20) cm = 30 cm = 0.3 m, k = 350/0.3 N/m

Substitute these values into equation 3

W = 1/2(350/0.3)(0.3²)

W = 350×0.3/2

W = 52.5 J

  the unit of the measure =

explanation:

  the property that is quantified by the energy needed per of material to raise the temperature by one.

  si unit of energy = joule (j)

  si unit of temperature = kelvin (k)

  si unit of mass of material = kilogram (kg)

the property that is quantified by the energy needed per of material to raise the temperature by one is called specific heat.

the unit of specific heat is