You drop a block from rest from a height h=3 m. The block lands on the spring at height h1 = 1.5 meters (spring constant = 160Nm) and compresses it 0.5 m before the block and spring momentarilty come to rest. (The spring then pushes the block upward.) What is the mass of the block?
If needed, g=10ms2.
I tried using kinematics (constant acceleration) for the first part of the motion (before the block hits the spring) and got v = 5.42218. Then I used conservation of energy for the block-spring system and got m = k*x^2/v^2 = 1.36054 but obviously it was wrong.
I just did one problem from you, the xo and vo sping
In total the block fell from 3 to 1 meters off floor
so change in gravitational potential
m g h = m g (2)
there it is stopped so that energy is now stored in the spring which compressed half a meter
m g (2) = (1/2) k x^2 = 80 (1/4) = 20
so
20 m = 20
m = 1 kg
riiight I forgot the gravitational potential energy. Thanks!
1.0 Kg
To find the mass of the block, we can use the principle of conservation of mechanical energy for the block-spring system.
Let's break down the problem into two parts:
1. The block falling from a height h=3m to the height h1=1.5m before hitting the spring.
2. The block compressing the spring by 0.5m and coming to rest momentarily.
Part 1: The block falling
Using the kinematic equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity (which is 0 since the block is dropped from rest), a is the acceleration (g), and s is the distance traveled, we can find the final velocity of the block just before it hits the spring.
v^2 = 0^2 + 2gh
v^2 = 2 * 10 * 3
v^2 = 60
v ≈ 7.74596 m/s
Part 2: The block compressing the spring
Using the principle of conservation of mechanical energy, we can equate the potential energy of the block at height h1 with the elastic potential energy stored in the compressed spring.
Potential energy of the block at height h1:
PE1 = mgh1
Elastic potential energy of the spring:
PE2 = (1/2)kx^2
As the block comes to rest momentarily, the potential energy is converted entirely into elastic potential energy:
mgh1 = (1/2)kx^2
m * 10 * 1.5 = (1/2) * 160 * 0.5^2
15m = 10
m = 10/15
m ≈ 0.66667 kg
Therefore, the mass of the block is approximately 0.66667 kg.