Question

In a room that is 2.41 m high, a spring (unstrained length = 0.30 m) hangs from the ceiling. A board whose length is 1.86 m is attached to the free end of the spring. The board hangs straight down, so that its 1.86-m length is perpendicular to the floor. The weight of the board (104 N) stretches the spring so that the lower end of the board just extends to, but does not touch, the floor. What is the spring constant of the spring?

Answer 1

Answer:

416 N\m

Explanation:

Weight of board

$W=ma\\\Rightarrow W=104\ N$

Length the spring can stretch = Total height of room - (unstrained length of spring + Length of Board)

$2.41-(0.3+1.86)\\\Rightarrow 2.41-2.16=0.25\ m$

From Hooke's law

$W=kx\\\Rightarrow k=\frac{W}{x}\\\Rightarrow k=\frac{104}{0.25}\\\Rightarrow k=416\ N\m$

The spring constant is 416 N/m