F on glider from spring = -kx
beginning x = .1 meter
potential energy in spring = kd^2/2 where d is extension
= 5 d^2/2 = 2.5 d^2
= 2.5 (.1-x)^2
the potential energy at start = 2.5(.1)^2 = .025 Joules total
the kinetic energy = (1/2)mv^2
= .1 (.2)^2 = .004 Joules
so
.025 = .004 + 2.5 (.1-x)^2
.0084 = (.1-x)^2
.0917 = .1-x
x = .00835 meter
check my algebra
A glider with mass m=0.200kg sits on a frictionless horizontal air track, connected to a spring with force constant k=5.00N/m. The glider is released from rest with the spring stretched 0.100 m.
What is the displacement x of the glider from its equilibrium position when its speed is 0.20 m/s?
1 answer
1 answer