ENERGY and MOMENTUM: How Fast Will the Block Move When a Compressed Spring is Released?

Q: A spring with a spring constant k = 100 N/m is compressed a distance ( x ) = 100 mm. A block with a mass ( m ) = 0.250 kg is placed next to the spring. The surface upon which the block rests is frictionless and horizontal. When the spring and block are released, how fast is the block moving as it leaves the spring?

A: The spring constant ( k ) is a measure of the force required to compress a spring by ( x ) meters ( m ). The SI unit of force is the Newton ( N ). Work will be required to compress the spring, and the amount of energy in Joules ( J ) needed to compress the spring is ( W ) = Fd. When the spring is fully compressed, it will possess a quantity of potential energy ( PE ) = ½ kx². When the spring is fully extended, the PE will have been converted into kinetic energy ( KE ). Kinetic energy is the “ energy of motion “, and its units are ½ mv². Our challenge is to determine the velocity ( v ) of the block as it leaves the spring. Velocity ( v ) is expressed in units of meters per second ( m/s ).

The law of conservation of energy states that energy cannot be created or destroyed-it only changes form. Thus, we can set PE equal to KE to determine the velocity of the block when it leaves the extended spring.

Note: Notice that the distance the spring is compressed ( x ) is expressed in millimeters. The SI unit of distance is the meter ( m ). Therefore, 100 mm = 0.100 m.

PE = KE

½ kx² = ½ mv²

kx² = mv² 

√ [ ( k/m ) ]( x ) = v

√ [ ( 100 N/m / 0.250 kg  ) ]( 0.100 m ) = v = 40 m/s

Q: Will the block continue moving at 40 m/s?

A: The surface upon which the block slides is frictionless; therefore, the force exerted by the spring is an unbalanced force. Force ( F ) = ma, where ( a ) is the acceleration of the block. The velocity of an object that accelerates under the influence of an unbalanced force is constantly increasing or decreasing. For this reason, acceleration may also be expressed as ( a ) = Δv/Δt. 

According to Hooke’s law, the force exerted on a compressed spring is F = kx. In this circumstance, the displacement ( x ) of the spring is in the same direction as the applied force. The force exerted by the spring opposes the force applied to it. In this circumstance, F_s = -kx.

F = kx = ma

a = ( k/m )( x )

a = ( 100 N/m / 0.250 kg )( 0.100 m ) = 40 m/s² 

The block’s velocity will increase by 40 m/s every second until its motion is altered by another force.