A 5.0-kg box is pulled by a horizontal force F applied to the top of the box. When the box meets a low doorstep, it begins to ro
1 answer:
Answer:
the required minimum magnitude of the force F is 21 N
Explanation:
Given the data in the question,
m = 5 kg
width = 60 cm
height = 80 cm
Let force is F represent in the image below,
so when the block about to rotate normal shifted to edge of cube
mg(w/2) = Fh
F = mg(w/2) / h
we know that g = 9.8 m/s²
we substitute
F = (5 × 9.8 ( 60/2)) / 70
F = (5 × 9.8 × 30 ) / 70
F = 1470 / 70
F = 21 N
Therefore, the required minimum magnitude of the force F is 21 N
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Answer:
a) perfectly inelastic, b) collision is inelastic, c) elastic
Explanation:
In this exercise, it is asked to identify what type of shock occurs between the skater and the frisbee, for this we must define a system formed by the skater and the fribee, so that the forces during the crash have been internal and the amount of movement is preserved
Initial instant. Before the skater touches the frisbee
p₀ = M v₁ + m v₂
where M and m are the masses of the skater and frisbee, respectively
for the final moment they give us several possibilities, in all case the moment is conserved
p₀ =
case a)
Final instant. grabs the frisbee and holds it
p_{f} = (M + m) v '
p₀ = p_{f}
We can see that this shock is perfectly inelastic, it holds the fressbee
case b)
final instant.
This case is similar to the previous one, but the final speed of fresbee is zero, therefore this collision is inelastic and the kinetic energy is not conserved.
case c)
final instant. Grab the fressbee and resend it
this is an elastic Shock since the equivalent of a rebound of the fressbee, the kinetic energy is conserved.