Question

physics genius solve this..... a coil spring fastenet at on end of force of 50 N streaches the spring 20centimeter find the work done ​

Answer 1

Work done is equals to force into displacement

20 cm is equals to 0.2 metre

wd= 0.2×50 N/m^2

=10N/m^2

Answer 2

Answer:

The spring constant is 500 N/m.

Explanation:

We use Hooke's Law, which states that,

F=kx

F is the force in newtons

k is the spring constant

x is the extension of the spring in meters

Solving for k, we get,

k=Fx

Now, we plug in the values, and we get,

k=50 N0.1 m

=500 N/m

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