Question

A spring with constant k 250N/m is extended by 3cm, how much force does the spring exert?

A. 833.3N
B. 83.33N
C. 75N
D. 75N

Answer 1

Using Hooke's Law, the force exerted by the spring with a constant k of 250 N/m extended by 3 cm is found to be 7.5 N.

Step-by-step explanation:

To determine the force exerted by a spring when it is extended, we use Hooke's Law which states that the force exerted by a spring (F) is directly proportional to the distance (x) it is stretched or compressed from its equilibrium position, mathematically expressed as F = kx. The spring constant (k) is given as 250 N/m, and the spring is extended by 3 cm (which is 0.03 m when converted to meters).

The calculation becomes:

• F = kx

• F = 250 N/m × 0.03 m

• F = 7.5 N

So, the force exerted by the spring is 7.5 N.