- #1
Abid Rizvi
- 20
- 0
Homework Statement
An inclined plane of angle θ = 20.0° has a spring of force constant k = 460 N/m fastened securely at the bottom so that the spring is parallel to the surface as shown in the figure below. A block of mass m = 2.51 kg is placed on the plane at a distance d = 0.324 m from the spring. From this position, the block is projected downward toward the spring with speed v = 0.750 m/s. By what distance is the spring compressed when the block momentarily comes to rest?
Homework Equations
K = 1/2 mv^2
Elastic Potential energy of spring = 1/2 kx^2
The Attempt at a Solution
So first I found the acceleration of the object:
gsin(θ) = a
Then I found the velocity of the mass at the time it hits the spring
d = vt +1/2 at^2
t is about .269593
so velocity when the object hits the spring = v+at which is about 1.654
Then I had: 1/2 mv^2 = 1/2 kx^2
solving for x, I got about 0.122 meters
However, the website to send the answer says I'm wrong. I have looked this up online, and I have substituted my values into other peoples equations and I still get 0.122
Thanks in advance!