A 50 g dart rests up against a spring that has been compressed 0.04 meters. It has a spring constant of 1560 N/m.What is the max
1 answer:
Answer:
The maximum velocity of the dart after the spring has transferred all its energy to it is 7.06 m/s.
Explanation:
Given that,
Mass of the dart, m = 50 g = 0.05 kg
Compression in the spring, x = 0.04 m
Spring constant, k = 1560 N/m
We need to find the maximum velocity of the dart after the spring has transferred all its energy to it. The kinetic energy is converted to elastic potential energy as :
So, the maximum velocity of the dart after the spring has transferred all its energy to it is 7.06 m/s.
You might be interested in
Answer:
a) see attached, a = g sin θ
b)
c) v = √(2gL (1-cos θ))
Explanation:
In the attached we can see the forces on the sphere, which are the attention of the bar that is perpendicular to the movement and the weight of the sphere that is vertical at all times. To solve this problem, a reference system is created with one axis parallel to the bar and the other perpendicular to the rod, the weight of decomposing in this reference system and the linear acceleration is given by
Wₓ = m a
W sin θ = m a
a = g sin θ
b) The diagram is the same, the only thing that changes is the angle that is less
θ' = 9/2 θ
c) At this point the weight and the force of the bar are in the same line of action, so that at linear acceleration it is zero, even when the pendulum has velocity v, so it follows its path.
The easiest way to find linear speed is to use conservation of energy
Highest point
Em₀ = mg h = mg L (1-cos tea)
Lowest point
Emf = K = ½ m v²
Em₀ = Emf
g L (1-cos θ) = v² / 2
v = √(2gL (1-cos θ))
