Question

The work done by an external force in stretching a spring having spring constant k= 100N/m by 10 cm from its unstretched position is ?

Answer 1

Answer:

0.5 Joule is the work done by an external force in stretching a spring having spring constant k= 100N/m by 10 cm from its unstretched position.

Explanation:

Given,

K = 100N/m

x = 10 cm

100 cm = 1 m

Then,

x = 10/100 m

x = 0.1 m

Ws = 1/2 K [(xi)^2 - (xf)^2]

Ws = 1/2 × 100 [(0)^2 - (0.1)^2] Joule

Ws = 50 [0 - 0.01] Joule

Ws = 50 × 0.01 Joule

Ws = 0.5 Joule

Here,

Ws is the spring force.

K is spring constant

x is the elongation or compression.

Hence,

0.5 Joule is the work done by an external force in stretching a spring having spring constant k= 100N/m by 10 cm from its unstretched position.

Answer 2

Answer:

1/2J. ,............. . . . .... .