please i really need help A spring has a force constant of 90000 N/m. How far mu
ID: 1970378 • Letter: P
Question
please i really need helpA spring has a force constant of 90000 N/m.
How far must it be stretched for its poten-
tial energy to be 30 J? Answer in units of m
An ore car of mass 37000 kg starts from rest
and rolls downhill on tracks from a mine. At
the end of the tracks, 28 m lower vertically, is
a horizontally situated spring with constant
3.3 × 105 N/m.
The acceleration of gravity is 9.8 m/s2 .
Ignore friction.
How much is the spring compressed in stop-
ping the ore car?
Answer in units of m
A block of mass m is pushed a distance D up
an inclined plane by a horizontal force F. The
plane is inclined at an angle with respect to
the horizontal. The block starts from rest and
the coefficient of kinetic friction is µk.If N is the normal force, the final speed of
the block is given by
Explanation / Answer
k = 90000 N/m
potential energy = (1/2)*K*x2
30 = 1/2 * 90000 x2
x= 0.026 m
change in potential energy of trolly = gain PE of spring
37000*9.8*28 = (1/2)(3.3*105)(x2)
x= 7.844 m
let the angle be
work cone by force = force* displacement along force
= F*Dcos
(mgsin is the force and N = mgcos...mg = N/cos)
work done by gravity =mg sin *D
= N /cos * sin *D
=N tan *D.........
work done by friction = k *N
work done by force = work done by gravity + work done by friction + KE
F*D cos = N tan *D + k *N + (1/2)*m*(v2)
v = [(2/m)(F*D cos - N tan *D - k *N)]
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