Answer:
Explanation:
Length if the bar is 1m=100cm
The tip of the bar serves as fulcrum
A force of 20N (upward) is applied at the tip of the other end. Then, the force is 100cm from the fulcrum
The crate lid is 2cm from the fulcrum, let the force (downward) acting on the crate be F.
Using moment
Sum of the moments of all forces about any point in the plane must be zero.
Let take moment about the fulcrum
100×20-F×2=0
2000-2F=0
2F=2000
Then, F=1000N
The force acting in the crate lid is 1000N
Option D is correct
Potential energy and work done are the same thing as much as kinetic energy and work done are the same thing. Potential energy is a state of the system, a way of storing energy as of virtue of its configuration or motion, while work done in most cases is a way of chaning this energy from one body to another.
I assume the 100 N force is a pulling force directed up the incline.
The net forces on the block acting parallel and perpendicular to the incline are
∑ F[para] = 100 N - F[friction] = 0
∑ F[perp] = F[normal] - mg cos(30°) = 0
The friction in this case is the maximum static friction - the block is held at rest by static friction, and a minimum 100 N force is required to get the block to start sliding up the incline.
Then
F[friction] = 100 N
F[normal] = mg cos(30°) = (10 kg) (9.8 m/s²) cos(30°) ≈ 84.9 N
If µ is the coefficient of static friction, then
F[friction] = µ F[normal]
⇒ µ = (100 N) / (84.9 N) ≈ 1.2
Answer:
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