That is true because if the object is moving at Forceful speeds than it will lose more of its kinetic energy
Answer:
the speed of the tip of a blade 10 s after the fan is turned off is 16.889 m/s.
Explanation:
Given;
diameter of the ceiling fan, d = 90 cm = 0.9 m
angular speed of the fan, ω = 64 rpm
time taken for the fan to stop, t = 28 s
The distance traveled by the ceiling fan when it comes to a stop is calculated as;

The speed of the tip of a blade 10 s after the fan is turned off is calculated as;

Therefore, the speed of the tip of a blade 10 s after the fan is turned off is 16.889 m/s.
Answer:
The angular velocity at the beginning of the interval is
.
Explanation:
Given that,
Angular acceleration 
Angular displacement 
Angular velocity 
We need to calculate the angular velocity at the beginning
Using formula of angular velocity


Where,
= angular acceleration
= angular velocity
Put the value into the formula



Hence, The angular velocity at the beginning of the interval is
.
Work is calculated by multiplying force by the distance that the object had moved. The applied force is 60 N, moving the object by 10 m. Thus, the work does is 600 J. For the friction force which is equal to,
100N x 0.250 = 25.0 N
the work done is,
W = (60 N - 25 N) x 10 m = 350 J
The kinetic energy of the box can be equated to this force. Thus, the answer is also 350 J.
Answer:
F = 156.3 N
Explanation:
Let's start with the top block, apply Newton's second law
F - fr = 0
F = fr
fr = 52.1 N
Now we can work with the bottom block
In this case we have two friction forces, one between the two blocks and the other between the block and the surface. In the exercise, indicate that the two friction coefficients are equal
we apply Newton's second law
Y axis
N - W₁ -W₂ = 0
N = W₁ + W₂
as the two blocks are identical
N = 2W
X axis
F - fr₁ - fr₂ = 0
F = fr₁ + fr₂
indicates that the lower block is moving below block 1, therefore the upper friction force is
fr₁ = 52.1 N
fr₁ = μ N
a
s the normal in the lower block of twice the friction force is
fr₂ = μ 2N
fr₂ = 2 μ N
fr₂ = 2 fr₁
we substitute
F = fr₁ + 2 fr₁
F = 3 fr₁
F = 3 52.1
F = 156.3 N