Explanation:
We will calculate the gravitational potential energy as follows.
=
= 1164000 J
or, = 1164 kJ (as 1 kJ = 1000 J)
Now, we will calculate the change in potential energy as follows.
=
=
= -873000 J
or, = -873 kJ
Thus, we can conclude that change in gravitational potential energy is -873 kJ.
What is the work done in stretching a spring by a distance of 0.5m if the restoring force is 48N is 12 J.
Hooke's Law states that the more you deform a spring, the more force it will take to deform it further.
Notations: F= Restoring force
k= spring constant
x = spring displacement
Given: F = 48N,
x= 0.5m
By using Formula; F= kx
F = 48 = kx, k = F/x
k = 48/0.5 = 96 N/m
work done by spring = 1/2 (kx)² = 1/2(96x0.5)² = 12 J
work done by spring is 12J.
Learn more about the Hooke's law with the help of the following link:
brainly.com/question/2449067
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<span>B) The period increases and the frequency decreases.
Hope this helps!
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Its Summer in the Argentina.
Correct option is B.
Answer:
(a) Magnitude of static friction force is 109 N
(b) Minimum possible value of static friction is 0.356
Solution:
As per the question;
Horizontal force exerted by the girl, F = 109 N
Mass of the crate, m = 31.2 kg
Now,
(a) To calculate the magnitude of static friction force:
Since, the crate is at rest, the forces on the crate are balanced and thus the horizontal force is equal to the frictional force, f:
F = f = 109 N
(b) The maximum possible force of friction between the floor and the crate is given by:
where
N = Normal reaction = mg
Thus
For the crate to remain at rest, The force exerted on the crate must be less than or equal to the maximum force of friction.