A spinning wheel
and the blade of a kitchen blender both illustrate rotation.
A child swinging on a swing
illustrates oscillation, or 'harmonic' motion.
A balloon being blown up is an example of dilation or inflation.
A sliding hockey puck demonstrates the concept of translation.
Force equals mass time acceleration. Weight is a force and it can replace force in the equation. The acceleration would be gravity, which is an acceleration.
1.)
Fw (weight) = m (mass) · g (gravity, 9.8 m/s²)
Fw = m * 9.81 m/s²
560N = m · 9.81 m/s²
m ≈ 57.08 kg
2.)
d = 350 meters
t = 65 seconds
velocity = d/t
velocity = 350 meters / 65 seconds
velocity ≈ 5.38 meters/sec
3.)
Force = 35N
Distance = 2 meters
Work = Force · Distance
Work = 35N · 2 meters
Work = 70 J
The force of static friction keeps a stationary object at rest. Once the force of static friction is overcome, the force of kinetic friction is what slows down a moving object.
Answer:
24 N
Explanation:
= mass of the cube = ![6.0 kg](https://tex.z-dn.net/?f=6.0%20kg)
Consider the three cubes together as one.
= mass of the three cubes together = ![3 m = 3 (6.0) = 18 kg](https://tex.z-dn.net/?f=3%20m%20%3D%203%20%286.0%29%20%3D%2018%20kg%20)
= acceleration of the combination = 2 ms⁻²
= Force applied on the combination
Using Newton's second law
![F = ma = (18) (2) = 36 N](https://tex.z-dn.net/?f=F%20%3D%20ma%20%3D%20%2818%29%20%282%29%20%3D%2036%20N)
= Force by the left cube on the middle cube
Consider the forces acting on left cube, from the force diagram, we have
![F - F_{L} = ma \\36 - F_{L} = (6) (2)\\F_{L} = 24 N](https://tex.z-dn.net/?f=F%20-%20F_%7BL%7D%20%3D%20ma%20%5C%5C36%20-%20F_%7BL%7D%20%3D%20%286%29%20%282%29%5C%5CF_%7BL%7D%20%3D%2024%20N)
Answer:
a) K = 0.63 J, b) h = 0.153 m
Explanation:
a) In this exercise we have a physical pendulum since the rod is a material object, the angular velocity is
w² =
where d is the distance from the pivot point to the center of mass and I is the moment of inertia.
The rod is a homogeneous body so its center of mass is at the geometric center of the rod.
d = L / 2
the moment of inertia of the rod is the moment of a rod supported at one end
I = ⅓ m L²
we substitute
w =
w =
w =
w = 4.427 rad / s
an oscillatory system is described by the expression
θ = θ₀ cos (wt + Φ)
the angular velocity is
w = dθ /dt
w = - θ₀ w sin (wt + Ф)
In this exercise, the kinetic energy is requested in the lowest position, in this position the energy is maximum. For this expression to be maximum, the sine function must be equal to ±1
In the exercise it is indicated that at the lowest point the angular velocity is
w = 4.0 rad / s
the kinetic energy is
K = ½ I w²
K = ½ (⅓ m L²) w²
K = 1/6 m L² w²
K = 1/6 0.42 0.75² 4.0²
K = 0.63 J
b) for this part let's use conservation of energy
starting point. Lowest point
Em₀ = K = ½ I w²
final point. Highest point
Em_f = U = m g h
energy is conserved
Em₀ = Em_f
½ I w² = m g h
½ (⅓ m L²) w² = m g h
h = 1/6 L² w² / g
h = 1/6 0.75² 4.0² / 9.8
h = 0.153 m