A 15.2-kg block rests on a horizontal table and is attached to one end of a (nearly) massless horizontal spring. By pulling horizontally on the other end of the spring, someone causes the block to accelerate uniformly and reach a speed of 5 m/s in 0.45 s. In the process, the spring is stretched by 0.2 m. The block is then pulled at a constant speed of 5 m/s, during which time the spring is stretched by only 0.05 m. Find the spring constant of the spring.
2007-01-26 18:23:24 · 3 answers · asked by Victor C 2 in Science & Mathematics ➔ Physics
If you're being asked a question like this, you've almost certainly been introduced to Hooke's Law.
F= kx
Where F = force
k = spring constant
x = distance
Also, remember Force equals mass times acceleration...
F = ma
Substituting, we get
ma = kx
m = 15.2, since that's the mass of the block and the spring is considered massless here.
a = This changes. For the first pull, we have 5m/s divided by .45s, so it's 11.11 meters per second squared.
You only need one pull to figure out the spring constant, so here we go.
ma = kx
m = 15.2
a = 11.11
k = spring constant
x = .2 (that's how far the block moved)
(15.2)(11.11) = (.2)k
You can work it out from there. Don't forget to use the right units!
This is what I remember from high school physics, anyway. If this is for homework, don't blame me if something goes wrong. I never had anyone do my homework for me...
2007-01-26 21:05:23 · answer #1 · answered by The Ry-Guy 5 · 0⤊ 0⤋
Just off the top of my head:
Since the acceleration is uniform, then the force on the block is constant. The work done by the force pulling on the spring goes partly into the KE of the block and the balance stored in the PE of the spring. You know the KE of the block, but not the PE in the spring.
Hey, where's friction?
For the constant velocity portion:
A constant speed implys balanced forces. This implys a frictional force which balance the force of the spring which would be k*(dx), where dx=0.05
This should allow using a Ff, friction force, to be put into the force balance of the first portion of the motion.
That's all I feel like doing now.
2007-01-26 18:47:09 · answer #2 · answered by modulo_function 7 · 0⤊ 0⤋
F=-kx
where F is spring force
k is spring constant
x is position (stretch)
also F=ma where a is accel and m is mass
therefore -kx=ma
-k=(ma)/x
m=15.2,
a=5/0.45
x=0.2
-k=844.4
That's all I could come up with.
The second part might involve the comparison of potential energy and force, but I can't find the spring equations.
2007-01-26 21:03:01 · answer #3 · answered by Labsci 7 · 0⤊ 0⤋