Question

A spring whose stiffness is 3500 N/m is used to launch a 4 kg block straight up in the classroom. The spring is initially compressed 0.2 m, and the block is initially at rest when it is released. When the block is 1.3 m above its starting position, what is its speed

Answer 1

Answer:

the speed of the block at the given position is 21.33 m/s.

Explanation:

Given;

spring constant, k = 3500 N/m

mass of the block, m = 4 kg

extension of the spring, x = 0.2 m

initial velocity of the block, u = 0

displacement of the block, d =1.3 m

The force applied to the block by the spring is calculated as;

F = ma = kx

where;

a is the acceleration of the block

$a = \frac{kx}{m} \\\\a = \frac{(3500) \times (0.2)}{4} \\\\a = 175 \ m/s^2$

The final velocity of the block at 1.3 m is calculated as;

v² = u² + 2ad

v² = 0 + 2ad

v² = 2ad

v = √2ad

v = √(2 x 175 x 1.3)

v = 21.33 m/s

Therefore, the speed of the block at the given position is 21.33 m/s.