Question

A spring of negligible mass and force constant 5 Nm^–1 is compressed by a distance x = 5 cm. A block of mass 200 g is free to leave the end of the spring. If the system is released, what will be the speed of the block when it leaves the spring?

Options

• 15 cms^–1

• 20 cms^–1

• 25 cms^–1

• 30 cms^–1

Answer 1

25 cms^–1 

Explanation:

Potential energy of the compressed spring is `1/2` kx². If v is the speed of the block, the PE of the spring is converted into the kinetic energy of the block `1/2` mv². Equating the two, we have `1/2` mv² = `1/2` kx²

∴ v = `x sqrt(k/m) = 5 xx 10^-2 xx sqrt(5/0.2)` = 25 cms^–1