Answer:
(E) μs(mA +mB)g
Explanation:
We can apply for mB:
∑ Fx = mB*a (→)
⇒ Ffriction = mB*a ⇒ a = Ffriction / mB = μs*N / mB
⇒ a = μs*(mB*g) / mB ⇒ a = μs*g (acceleration of the system)
Now, for mA we have
∑ Fx = mA*a (→)
F - Ffriction = mA*a ⇒ F = mA*a + Ffriction
⇒ F = mA*(μs*g) + μs*(mB*g) ⇒ F = μs*g*(mA + mB)
We must know that the friction acts only between the two blocks
The correct answer is C) towards the center of the circle.
Although the object is moving at a constant speed it is constantly accelerating due to the constant change in direction as it describes the circular path. This causes a constant change in velocity as velocity is a vector quantity.
For the object to maintain the circular path there has to be centripetal force acting on the object and this centripetal force is directed towards the center of the circle.
It’s either 0.05 or 20. Assuming that the coefficient friction is a damping factor, I feel like 0.05 would be correct m
Oh my lord lol I was do ready to help then I saw numbers
