How much work must be done on a spring (k = 730 n/m) to stretch is by 1.5 m?

The work which should be done on the spring to stretch it by 1.5 m us 547.5 N.

Explanation

The spring has elastic property of getting deformed on stretching and attaining the original shape once, the stretching force is removed.

So according to Hooke’s law, the stress applied on the spring will be directly proportional to the strain or stretching occurred in the spring.

So according to Hooke’s law,

F=k.x

Here k is the spring constant and x is the distance of stretching.

As the spring is stretched in one direction only, the spring constant will be k/2.

So, the work done can be calculated from,

W=F.x=k/2.x^2

So on substituting the values of k as 730 N/m and stretching distance x as 1.5 m, we will get

W=730/2*1.5=547.5 N

Thus, the work which should be done on the spring to stretch it by 1.5 m us 547.5 N.

The work must be equal to 1642.5J

Explanation: Work is calculated as the movement caused by a force, so the work can be written as:

W = F*d

Where F is the force and d is the distance moved.

Now, the force in a spring is written as:

F = k*x where k is the spring constant and x is the distance stretched, in this case, x = d.

So we have that:

W = F*d = k*d^2

now, k = 730N/m and d = 1.5m

W =  730N/m*(1.5m)^2 = 1642.5 N*m = 1642.5J

if they have a finite resultant

the components of a vector are the parts of the vectors that are to each other.

b perpendicular