A box weighing 200N is pushed on a horizontal floor. What acceleration will result if a horizontal force of 100N is applied on t
2 answers:
Answer:
The acceleration of the block is 0.98 m/s².
Explanation:
It is given that,
Weight of a box, W = 200 N
Horizontal force acting on the box, F = 100 N
Coefficient of sliding friction is 0.4
Net force acting on the box,
Let a is the acceleration of the box. So,
Since,
So, the acceleration of the block is 0.98 m/s². Hence, this is the required solution.
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Answer:
Answer is explained in the explanation section.
Explanation:
A)
Solution:
For this to find, we need to calculate the centripetal acceleration on the equator.
The centripetal acceleration of the equator:
a = 4RcosФ/
,
where,
R is the radius of the earth
R = 6378 KM = 6.3 x and
T is the time period
T = 24 h = 86164.1 s
At Equator, Ф = 0°
So, CosФ = 1
Hence,
a = 4R/
By plugging in the values, we get:
a = 4 x () x (6.3 x
) /
a = 0.03 m/
Hence, this is the centripetal acceleration on the equator. And we also know that, acceleration due to gravity is 9.8 m/ which is very higher than the centripetal acceleration on the equator.
B) Normal force exerted by chair will always be equal and opposite to the mass times gravitational acceleration (F = mg). Otherwise, I would be thrown away from chair in case the normal force is not equal and opposite or I would be drag down to the earth due to greater mass times gravitational acceleration. Hence, both are equal and opposite.
C) Of course, this is not a lie, it is true because the acceleration due to gravity is 9.8 m/ and as we calculated the acceleration on the equator is 0.03 m/
which way too low to experience.
For percentage difference,
9.8 - 0.03 = 9.77
So, % diff = (9.8 - 9.77)/9.8 x 100
% diff = 0.00306 x 100
% diff = 0.306%
Obviously, this is way too low to experience.
D) With the help of same formula as discussed above, we have:
a = 4RcosФ/
,
Here, Ф = 44.4°
Just putting the values. we get
a = 0.0241 m/
Acceleration while sitting in my chair.