A 0.230-kg block along a horizontal track has a speed of 1.70 m/s immediately before colliding with a light spring of force constant 48.0 N/m located at the end of the track. What is the spring's maximum compression if the track is frictionless?
mv²/2=kx²/2
x=v•sqrt(m/k) =1.7•sqrt(0.23/48)=0.12 m
To find the spring's maximum compression, we need to apply the principles of conservation of energy and Hooke's Law.
First, let's calculate the initial kinetic energy of the block before the collision. The equation for kinetic energy is given by:
KE_initial = (1/2) * m * v^2
where
m = mass of the block = 0.230 kg
v = speed of the block = 1.70 m/s
Plugging in the given values, we have:
KE_initial = (1/2) * 0.230 kg * (1.70 m/s)^2
Next, let's consider the spring's potential energy at maximum compression. According to Hooke's Law, the potential energy stored in a spring is given by:
PE_spring = (1/2) * k * x^2
where
k = force constant of the spring = 48.0 N/m (given)
x = maximum compression of the spring (what we need to find)
Since the block comes to a momentary stop at maximum compression, all its initial kinetic energy is converted into potential energy stored in the spring. Therefore, we can equate the initial kinetic energy to the potential energy:
KE_initial = PE_spring
(1/2) * 0.230 kg * (1.70 m/s)^2 = (1/2) * 48.0 N/m * x^2
Simplifying the equation, we get:
(0.230 kg * 1.70 m/s^2) = (48.0 N/m) * x^2
0.391 kg m/s^2 = 48.0 N/m * x^2
Now, let's solve for x:
x^2 = (0.391 kg m/s^2) / (48.0 N/m)
x^2 = 0.0081458333 m^2
Taking the square root of both sides, we find:
x = √(0.0081458333 m^2)
x ≈ 0.0903 m
Therefore, the spring's maximum compression is approximately 0.0903 meters.