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To solve this problem we will apply the principle of conservation of energy. For this purpose, potential energy is equivalent to kinetic energy, and this clearly depends on the position of the body. In turn, we also note that the height traveled is twice that of the rigid rod, therefore applying these concepts we will have
Therefore the minimum speed at the bottom is required to make the ball go over the top of the circle is 4.67m/s
The force of frictions is opposed to relative motion.
The acceleration of the crate is approximately <u>2.937 m/s²</u>.
Reason:
The given parameters are;
The mass of the wood, m = 100 kg
The force which can move the wood, F = 588 N
Wood on wood static friction, = 0.5
Wood on wood kinetic friction, = 0.3
Solution;
The force of friction, , acting when the crate is moving is given as
follows;
= m × g ×
Where;
g = The acceleration due to gravity ≈ 9.81 m/s²
Therefore, we have;
= 100 × 9.81 × 0.3 = 294.3
The force of friction, = 294.3 N
The force with which the crate moves, F = 588 - 294.3 = 293.7
The force with which the crate moves, F = 293.7 N
Force = Mass, m × Acceleration, a
Therefore;
The acceleration of the crate, a ≈ <u>2.937 m/s²</u>.
Learn more about friction here: