Question

A spring with load 5 kg is stretched by 40 cm. Determine its spring constant.

Answer 1

Answer:

37.5 Nm

Explanation:formula is f=-kx

Answer 2

Answer:

The value of the spring constant is 125N/m.

Explanation:

We calculate the force(F) in two ways,

$F=mg$      (1)

$F=kx$       (2)

Where,

m=mass of the body

g=acceleration due to gravity=10m/s²

k=spring constant

x=distance through which the spring is stretched

From the question we have,

Mass of the load(m)=5 kg

The distance through which the spring is stretched=40cm=0.4m

We can derive the following result by combining equations (1) and (2):

$mg=kx$   (3)

We get the following by inserting the numbers in equation (3):

$5\times 10=k\times 0.4$

$k=\frac{50}{0.4}$

$k=125\ N/m$

Hence, the value of the spring constant is 125N/m.

#SPJ3