Refer to the diagram shown below.
m = the mass of the object
x = the distance of the object from the equilibrium position at time t.
v = the velocity of the object at time t
a = the acceleration of the object at time t
A = the amplitude ( the maximum distance) of the mass from the equilibrium
position
The oscillatory motion of the object (without damping) is given by
x(t) = A sin(ωt)
where
ω = the circular frequency of the motion
T = the period of the motion so that ω = (2π)/T
The velocity and acceleration are respectively
v(t) = ωA cos(ωt)
a(t) = -ω²A sin(ωt)
In the equilibrium position,
x is zero;
v is maximum;
a is zero.
At the farthest distance (A) from the equilibrium position,
x is maximum;
v is zero;
a is zero.
In the graphs shown, it is assumed (for illustrative purposes) that
A = 1 and T = 1.
Mechanical energy = potential energy + kinetic energy
The ball is on the ground so it has no potential energy. that's all i know.
Answer:
250 m/s
Explanation:
The mass of the bullet, m₁ = 100 g = 0.1 kg
The mass of the gun, m₂ = 5 kg
The backward velocity of the gun, v₂ = -5 m/s
Given that the momentum is conserved, we have;
The total initial momentum = The total final momentum
The gun and the bullet are at rest, therefore, we have;
The initial momentum = 0
The total final momentum = m₁·v₁ + m₂·v₂
Where;
v₁ = The forward velocity of the bullet
Therefore, we get;
m₁·v₁ + m₂·v₂ = 0
0.1 kg × v₁ + 5 kg × (-5 m/s) = 0
0.1 kg × v₁ = 5 kg × 5 m/s
v₁ = (5 kg × 5 m/s)/(0.1 kg) = 250 m/s
The forward velocity of the bullet, v₁ = 250 m/s
Answer:
2,352 Joules
Explanation:
At the ground, the barbell has a classical mechanical energy value of zero. There is no classical kinetic or potential energy for the barbell. The moment the man starts to lift the barbell, he does work on the barbell and transfers kinetic energy to it due to the motion. At its maximum height where the man lifts the barbell to a stop, the kinetic energy is zero because it transformed into gravitational potential energy stored in the gravitational field. Our reference point for potential was defined to be zero at the floor, therefore we can say that the gravitational potential energy at 2 meters is:

(2) impulse is a vector quantity. Energy has no direction, nor does power. Work is a dot product of vector quantities, which makes it a scalar.