Answer:
350 N/m
Explanation:
If we are assuming the stretch does not exceed the elastic range of the material, then by Hooke's law the spring constant of the cord is simply the ratio between the force 70N acting on the cord to stretch 20cm or 0.2m
k = 70 / 0.2 = 350 N/m
The spring constant is 350 N/m
The car's mass is 1600 kg.
Its weight is (mass) x (gravity).
On Earth, that's (1600 kg) x (9.8 m/s²) = 15,680 Newtons.
At the moment, that's the only force acting on the car, directed downward and provided by gravity.
If you want to lift the car, then the net force has to be directed upward, and must either exactly cancel or exceed the force of gravity.
So the minimum force required to lift the car is <em>15,680 Newtons</em>, directed vertically upward.
Answer:
The magnitude of acceleration is reduced.
Explanation:
Force is defined as push or pull
The force is said to be<em> balance force </em>if the force are equal in size but opposite in direction. ie the object does not move or move with constant speed.
The force are to be<em> unbalanced force </em>if the force cause change in motion. ie the object has force greater than zero and has acceleration.
According to <em>Newton second law of motion </em>, acceleration depends on force acting on the object and mass of object.
F=ma
a=
When unbalanced force act on the mass of object it reduces magnitude of acceleration without changing the direction.
Answer:
a)
= 928 J
, b)U = -62.7 J
, c) K = 0
, d) Y = 11.0367 m, e) v = 15.23 m / s
Explanation:
To solve this exercise we will use the concepts of mechanical energy.
a) The elastic potential energy is
= ½ k x²
= ½ 2900 0.80²
= 928 J
b) place the origin at the point of the uncompressed spring, the spider's potential energy
U = m h and
U = 8 9.8 (-0.80)
U = -62.7 J
c) Before releasing the spring the spider is still, so its true speed and therefore the kinetic energy also
K = ½ m v²
K = 0
d) write the energy at two points, maximum compression and maximum height
Em₀ = ke = ½ m x²
= mg y
Emo = 
½ k x² = m g y
y = ½ k x² / m g
y = ½ 2900 0.8² / (8 9.8)
y = 11.8367 m
As zero was placed for the spring without stretching the height from that reference is
Y = y- 0.80
Y = 11.8367 -0.80
Y = 11.0367 m
Bonus
Energy for maximum compression and uncompressed spring
Emo = ½ k x² = 928 J
= ½ m v²
Emo =
Emo = ½ m v²
v =√ 2Emo / m
v = √ (2 928/8)
v = 15.23 m / s