Question

A spring is mounted horizontally as shown to the right. A crate,
which has a mass of 8.5 kg is pressed against the spring with a
force of 350 N. As a result the spring is compressed a distance of
82.0 cm. The mass is then released and is allowed to slide along
the horizontal, frictionless surface.
a. What is the spring constant of this spring?
b. How much elastic potential energy will be stored in the
spring?
c. How much work will be done in compressing this spring?
d. What will be the kinetic energy of this crate after it has left the spring?
e. What will be the velocity of the crate after it has left the spring?

Answer 1

a.)

Force on a spring is given by,

F = k*dx

given, F = 350 N

k = spring constant = ??

dx = 82.0 cm = 0.82 m

So, k = F/dx = 350/0.82

k = 426.83 N/m

b.)

Potential energy stored in spring is given by,

E = 0.5*k*x^2

from known values:

E = 0.5*426.83*0.82^2

E = 143.5 J

c.)

Since work done is given by,

W = change in spring energy = dE = Ef - Ei

Ei = initial spring energy = 0

Ef = final spring energy = 143.5 J

W = 143.5 - 0

W = 143.5 J

d.)

from energy conservation,

W = dKE = change in kinetic energy

W = KEf - KEi

here, KEi = initial kinetic energy = 0

KEf = final kinetic energy = ??

So,

143.5 = KEf - 0

KEf = 143.5 J

e.)

noe, kinetic energy is given by,

KE = 0.5*m*v^2

so, finally,

KEf = 0.5*m*Vf^2

given, m = mass = 8.5 kg

Vf = velocity of the crate after it has left the spring = ??

So,

143.5 = 0.5*8.5*Vf^2

Vf = sqrt(2*143.5/8.5)

Vf = 5.81 m/sec.

Let me know if you've any query.