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Newtonian mechanics discusses the movement of classical bodies under the influence of forces by applying Newton’s three laws. For more general concepts, use [classical-mechanics]. For Newton’s description of gravity, use [newtonian-gravity].
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Amplitude of Shm if Constant External force is applied
The addition of the constant F will shift the equilibrium position to the left. Then the "right hand amplitude" and the "left hand amplitude" is the same with respect to this new equilibrium position. …
2
votes
Combination of perpendicular simple harmonic motions
First, I am interpreting your "phase difference" as the phase difference between the x-direction and y-direction oscillation.
Think about the two limiting cases: if the phase difference is zero, the …
0
votes
Accepted
Pulling yourself towards an object by a rope
Welcome to Physics StackExchange!
If both you and the sledge are on a frictionless surface, and the rope is massless, then yes, your application of Newton's third law is correct, you will accelerate …
1
vote
Accepted
Getting different answers from different calculations of a simple mechanics problem and can'...
You are absolutely right in stating that the “added energy comes from the explosion”. In order to use conservation of energy, the initial energy calculation need to captures all forms of energy presen …
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vote
Accepted
What is the physical condition of a loose pendulum string?
Your assumption is indeed wrong. Hint: the pendulum is undergoing circular motion up until the point when the thread is not fully extended.
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Accepted
What happens if you spin a coin in space?
Assume there is no other force (such as friction), and say the coin is near the Earth and experiences the Earth's gravitational field, then with classical mechanics, the gravitational force only acts …
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Why we say that total energy should be zero at infinity in the derivation of escape velocity?
Your mistake is claiming that there is a gravitational force at infinity, when in fact, it is precisely zero. You can see that from the limit
$$
\lim_{r\rightarrow \infty} G\frac{Mm}{r^2} = 0
$$
So an …