Question

A 1.70 kg block slides on a horizontal, frictionless surface until it encounters a spring with a force constant of 955 N/m. The block comes to rest after compressing the spring by a distance of 4.60 cm. The other end of the spring is attached to a wall. Find the initial speed of the block.

Answer 1

Answer:

The initial speed of the block is 1.09 m/s

Explanation:

Given;

mass of block, m = 1.7 kg

force constant of the spring, k = 955 N/m

compression of the spring, x = 4.6 cm = 0.046 m

From principle of conservation of energy

kinetic energy of the block = elastic potential energy of the spring

¹/₂mv² = ¹/₂kx²

mv²  = kx²

$v = \sqrt{\frac{kx^2}{m} }$

where;

v is the initial speed of the block

x is the compression of the spring

$v = \sqrt{\frac{955*(0.046)^2}{1.7} } \\\\v = 1.09 \ m/s$

Therefore, the initial speed of the block is 1.09 m/s