Springs physics problem?
A 0.266-kg block is dropped straight downwards onto a vertical spring. The spring constant of the spring is 61 N/m. The block sticks to the spring and the spring compresses 0.16 m before coming to a momentary halt. What is the speed of the block just before it hits the spring?
Comments
m = mass of the block = 0.266 kg
k = spring constant = 61 N/m
d = spring compression = 0.16 m
g = acceleration by gravity = 9.8 m/s²
v = speed of the block at the moment of impact = ?
The moment the block touches the spring, it has some speed, giving it some kinetic energy.
At that moment, it still is some distance above its lowest point, giving it also some gravitational potential energy.
The total elastic potential energy stored in the spring at its maximum compression will be the sum of that kinetic energy and gravitational potential energy.
Kinetic energy of the block:
KE = m×v²/2
KE = (0.266 kg)×v²/2
KE = (0.133 kg)×v²
Gravitational potential energy of the block:
GPE = m×g×d
GPE = (0.266 kg)×(9.8 m/s²)×(0.16 m )
GPE = 0.417088 J
Elastic potential energy in the spring:
EPE = k×d²/2
EPE = (61 N/m)×(0.16 m)²/2
EPE = 0.7808 J
By conservation of energy:
KE + GPE = EPE
(0.133 kg)×v² + (0.417088 J) = (0.7808 J)
v = 1.65 m/s < - - - - - - - - - - - - - - - - - - - - - - - - answer
GPE of block lost as spring is compressed = mgx = (0.266)(9.81)(0.16) = 0.418 J
The SPE of compressed spring = 1/2kx² = (0.5)(61)(0.16)² = 0.781 J
The total mechanical energy (ME) of block as it hits spring is:
ME = GPE + SPE = 0.418 + 0.781 = 1.199 J
KE of block as it hits spring = ME = 1.199 = 1/2mV² = (0.5)(0.266)V² = 0.133V²
V² = 1.199/0.133 = 9.01504
V = 3.00 m/s ANS