Question

A spring of negligible mass has ks = 1600N/m. It is placed horizontally with one end fastened to a wall.
a.) How far must the spring be compressed for 3.50J of potential energy to be stored in it?
b.) A 1.20kg ball is rolled down a ramp whose top is 0.80m above the spring and reaches the horizontal just in front of the spring. What is the kinetic energy of the ball just before it contacts the spring?
c.) What is the velocity of the ball at that point using energy methods?
d.) After the ball contacts the spring, what distance will the spring be compressed from its equilibrium point?

Answer 1

Actually, believe I have part a

Answer 2

Just no clue for the rest

Answer 3

For part a, the formula for elastic energy is 1/2 kx^2, where k is the spring constant, and x is the distance the spring is stretched. So for
Ek=3.5
kx^2=7
x^2=.004375
x=.066 meters, or 6.6 centimeters

For part b, we know that gravitational energy at the top is going to equal the kinetic energy just before it hits the spring.
Eg=gravity*mass*height
Eg=9.8*1.2*.8
Eg= 9.408 J

For part c, we know that the gravitational energy for the ball at the top of the ramp is all going to be converted into kinetic energy. So
Eg=Ek
gravity*mass*height= 1/2 *mass*velocity^2
9.8*1.2*.8=1/2*1.2*v^2
v^2=15.68
v=3.96 m/s

For part d, when the spring is completely compressed, all of the energy will be elastic. So, we can set the total energy equal to elastic.
9.408=1/2 *constant*distance^2
18.816=1600*distance^2
.01176=distance^2
distance=.108 meters, or 10.8 centimeters

Answer 4

Thanks, your a god send