Answer:
a) 500 N
b) 250 J
c) 0.87 m
d) 1.58 m/s, at 0.6 m from the wall
Explanation:
The mass of the object m = 100 kg
the spring constant k = 1000 N/m
length of the the spring = 1 m
extension of the string = 50 cm = 0.5 m
a) Force used to pull the mass is gotten from Hooke's law equation
F = -kx
where F is the force used to pull = ?
k is the spring constant = 1000 N/m
x is the extension = 0.5 m
substituting, we have
F = 1000 x 0.5 = 500 N this force is used to pull the mass
b) The work done in moving the mass = Fx
==> 500 x 0.5 = 250 J
c) The energy stored up in the spring U = [tex]\frac{1}{2}kx^{2}[/tex]
U = [tex]\frac{1}{2}*1000*0.5^{2}[/tex] = 125 J
energy available for the mass from its equilibrium position = 250 - 125 = 125 J
this energy is equivalent to the work done by the spring on the mass by moving it closer to the wall
Work W = (weigh of the mass) x distance moved
weight = mg
where m is the mass = 100 kg
g is acceleration due to gravity = 9.81 m/s^2
substituting, we have
W = mgd
where d is the distance the mass moves closer to the wall
W = 100 x 9.81 x d
but W = 125 J
125 = 981d
d = 125/981 = 0.13 m
closeness to the wall = L - d
where L is the natural length of the spring = 1 m
closeness to the wall = 1 - 0.13 = 0.87 m
d) The maximum kinetic energy of the object will be halfway between the extended length and the final resting place.
extended length = 1 + 0.5 m = 1.5 m
distance from resting place = 1.5 - 0.87 = 0.63 m from the wall
At this point, all the mechanical energy on the mass and spring system is converted to kinetic energy of motion.
KE = [tex]\frac{1}{2}mv^{2}[/tex]
substituting,
125 = [tex]\frac{1}{2}*100*v^{2}[/tex] = [tex]50v^{2}[/tex]
[tex]v^{2}[/tex] = 125/50 = 2.5
v = [tex]\sqrt{2.5}[/tex] = 1.58 m/s
