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

The initial speed of the block is 1.09 m/s

Step-by-step 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{(kx^2)/(m) }`

where;

v is the initial speed of the block

x is the compression of the spring

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

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