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
You might be interested in
Answer:
-54.12 V
Explanation:
The work done by this force is equal to the difference between the final value and the initial value of the energy. Since the charge starts from the rest its initial kinetic energy is zero.
The change in electrostatic potential energy , of one point charge q is defined as the product of the charge and the potential difference.
The efficiency of an ideal Carnot heat engine can be written as:
where
is the temperature of the cold region
is the temperature of the hot region
For the engine in our problem, we have
and 
, so the efficiency is
where
For the engine in our problem, we have