Initially sliding with a speed of 2.5 m/s, a 2.0 kg block collides with a spring and compresses it 0.35 m before coming to rest. What is the force constant of the spring?
2 answers
K= N/M
Two equations:
KE=1/2mv^2
and
PE=1/2kx^2
Where:
PE=potential energy
KE=Kinetic energy
m=mass of the block
v=velocity of the block
k=spring constant
and
x=0m-0.35m=-0.35m
Set both equations equal together and solve for k:
1/2mv^2=-1/2kx^2
k=mv^2/x^2
k=[2.0kg*(2.5m/s)^2]/(0.35m)^2
Answer should contain no more than two significant figures.
KE=1/2mv^2
and
PE=1/2kx^2
Where:
PE=potential energy
KE=Kinetic energy
m=mass of the block
v=velocity of the block
k=spring constant
and
x=0m-0.35m=-0.35m
Set both equations equal together and solve for k:
1/2mv^2=-1/2kx^2
k=mv^2/x^2
k=[2.0kg*(2.5m/s)^2]/(0.35m)^2
Answer should contain no more than two significant figures.