A mass of 10 kg is at a point A on the table. It moved to a point B. If the line joining A and B is horizontal. What is the work
1 answer:
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Answer:
option C
Explanation:
given,
diameter of circular room = 8 m
rotational velocity of the rider = 45 rev/min
=
=4.712 rad/s
here in this case normal force is equal to centripetal force
N = m r ω²
N = m x 4 x 4.712²
N = 88.83m
frictional force = μ N
= 88.83m x μ
now, for the body to not to slide
gravity force is equal to frictional force
m g = 88.83 m x μ
g = 88.83 x μ
9.8 = 88.83 x μ
μ = 0.11
hence, the correct answer is option C
Answer:
The x-component of is 56.148 newtons.
Explanation:
From 1st and 2nd Newton's Law we know that a system is at rest when net acceleration is zero. Then, the vectorial sum of the three forces must be equal to zero. That is:
(1)
Where:
,
,
- External forces exerted on the ring, measured in newtons.
- Vector zero, measured in newtons.
If we know that ,
,
and
, then we construct the following system of linear equations:
(2)
(3)
The solution of this system is:
,
The x-component of is 56.148 newtons.