The mass and spring are resting on a horizontal tabletop. The spring constant is 50 N/m, and the mass is stretched to 48cm. It is then released and the spring-mass system begins to oscillate. Predict the speed of the mass as it passes a point that is 35cm from its equilibrium point on the other side of the equilibrium position (spring is compressed) using an algebraic expression of energy conservation.
Help!
Help!
-
Use the Total Mechanical Energy equations for a spring and conservation of energy to solve:
E=1/2mv^2+1/2kx^2
Initially, the spring is stretched to 48cm(0.48m) and is at rest, so it has no kinetic energy. So, we solve for the potential energy, since, at this time, it is equal to the total energy of the spring:
E=1/2kx^2
k=50 N/m
x=0.48 m
E=1/2(50)(0.48)^2
E=5.76 J
When the mass is at 35cm(0.35m) the spring has both kinetic and potential energy which together equal the total energy we just solved for, since energy is conserved. So, we solve for v in the total energy equation:
E=1/2mv^2+1/2kx^2
E=5.76 J
m=0.5 kg
k=50N/m
x=0.35 m
5.76=1/2(0.5)v^2+1/2(50)(0.35)^2
5.76=0.25v^2+3.0625
0.25v^2=2.6975
v^2=10.79
v=3.285m/s
E=1/2mv^2+1/2kx^2
Initially, the spring is stretched to 48cm(0.48m) and is at rest, so it has no kinetic energy. So, we solve for the potential energy, since, at this time, it is equal to the total energy of the spring:
E=1/2kx^2
k=50 N/m
x=0.48 m
E=1/2(50)(0.48)^2
E=5.76 J
When the mass is at 35cm(0.35m) the spring has both kinetic and potential energy which together equal the total energy we just solved for, since energy is conserved. So, we solve for v in the total energy equation:
E=1/2mv^2+1/2kx^2
E=5.76 J
m=0.5 kg
k=50N/m
x=0.35 m
5.76=1/2(0.5)v^2+1/2(50)(0.35)^2
5.76=0.25v^2+3.0625
0.25v^2=2.6975
v^2=10.79
v=3.285m/s
-
Potential energy in spring = 0.5 k x^2
kinetic energy in mass = 0.5 mv^2
max energy in spring = 0.5(50) 0.48^2
energy in spring at 35 cm = 0.5(50) (0.35)^2
the difference between the max energy and energy at 35cm is conserved it goes into Kinetic energy
0.5(50) 0.48^2 - 0.5(50) (0.35)^2 = 0.5 mv^2
50( 0.48^2 - 0.35^2) = mv^2
5.395 = mv^2
v= sqrt(5.395/m)
kinetic energy in mass = 0.5 mv^2
max energy in spring = 0.5(50) 0.48^2
energy in spring at 35 cm = 0.5(50) (0.35)^2
the difference between the max energy and energy at 35cm is conserved it goes into Kinetic energy
0.5(50) 0.48^2 - 0.5(50) (0.35)^2 = 0.5 mv^2
50( 0.48^2 - 0.35^2) = mv^2
5.395 = mv^2
v= sqrt(5.395/m)