A 5.00×105-kg subway train is brought to a stop from a speed of
0.500 m/s in 0.400 m by a large spring bumper at the end of its track.
What is the force constant k of the spring?
mvi^2=kxf
k=mvi^2 / xf2
k=(5.00*105)(0.5)^2 / (0.4)
k=328 N/m
oops, forgot to square my x
To find the force constant (k) of the spring, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement it undergoes.
Hooke's Law: F = -kx
Where:
F = Force exerted by the spring
k = Force constant of the spring
x = Displacement of the spring
In this case, the subway train is being brought to a stop, so it experiences a deceleration. The opposing force responsible for this deceleration is the force exerted by the spring bumper. The magnitude of this force is given by:
F = m * a
Where:
m = Mass of the subway train
a = Acceleration
The acceleration can be calculated using the kinematic equation:
v^2 = u^2 + 2as
Where:
v = Final velocity (0 m/s since the subway train comes to a stop)
u = Initial velocity (0.500 m/s)
a = Acceleration
s = Displacement (0.4 m)
Rearranging the equation to solve for a:
a = (v^2 - u^2) / (2s)
Now we can substitute the given values and solve for the acceleration:
a = (0 - 0.500^2) / (2 * 0.400)
a = -0.0625 m/s^2 (negative sign indicates deceleration)
Next, we can calculate the force exerted by the spring using:
F = m * a
Substituting the given mass of the subway train:
F = 5.00 × 10^5 kg * (-0.0625 m/s^2)
F = -31250 N (negative sign indicates the force is in the opposite direction of motion)
Finally, we can use Hooke's Law to find the force constant (k). Rearranging the equation:
k = -F / x
Substituting the values:
k = -(-31250 N) / 0.400 m
k = 78125 N/m
Therefore, the force constant (k) of the spring is 78125 N/m.
KE=PE
mv²/ 2 =kx²/2
k=m(v/x) ²