Answer:
U₁ = 129.4 J
Explanation:
The potential energy is
U = mg y - m g y₀
Where I correspond to the initial position, with this it is an additive constant, we can make it zero with the placement of the reference system, in this case the system is placed on the floor where the ladder rests.
The power power for people on the floor is
U₀ = 0 J
The potential energy for the person on the first step is
U₁ = m g y₁
In general the steps are 20 cm high
y₁ = 20 cm (1m / 100cm) = 0.20 m
U₁ = 66 9.8 0.20
U₁ = 129.4 J
The answer is 4.0 kg since the flywheel comes to rest the
kinetic energy of the wheel in motion is spent doing the work. Using the
formula KE = (1/2) I w².
Given the following:
I = the moment of inertia about the
axis passing through the center of the wheel; w = angular velocity ; for the
solid disk as I = mr² / 2 so KE = (1/4) mr²w². Now initially, the wheel is spinning
at 500 rpm so w = 500 * (2*pi / 60) rad / sec = 52.36 rad / sec.
The radius = 1.2 m and KE = 3900 J
3900 J = (1/4) m (1.2)² (52.36)²
m = 3900 J / (0.25) (1.2)² (52.36)²
m = 3.95151 ≈ 4.00 kg
Answer:
Explanation:
If F is the applied force and θ is the applied angle from horizontal with positive angle below the horizon
Fnet = Fcosθ - μ(mg + Fsinθ)
The only variable you've actually given us is the friction coefficient.
You'll have to plug your own numbers in.
Remember, if the applied force is acting towards the floor, the normal force is increased as will be the friction force. The net force will decrease.
If the applied force is acting upward, the θ angle will be negative and the normal force is decreased along with the friction force. The net force will increase.
Answer:
Explanation:
Given
mass of sled =26 kg
coefficient of static friction
coefficient of kinetic friction
In order to move sled from rest we need to provide a force greater than static friction which is given by
After Moving Sled kinetic friction comes in to play which is less than static friction
therefore minimum force to keep moving sledge at constant velocity is 18.34 N