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14

The confusion comes from how you have written the equation. If you write it like this $$F_{net} = ma$$ it will be easier to see your error. You are exerting a force on the boulder, but net force is zero. This means that other forces such as friction are canceling your force out. In this case. Friction equals applied force.


3

If $s_0$ and $u$ are zero, then the equaion for $s$ simplifies to: $$ s = \tfrac{1}{2}at^2 $$ so: $$ \frac{ms}{t^2} = \tfrac{1}{2}ma $$ and since $F=ma$ this becomes: $$ \frac{ms}{t^2} = \tfrac{1}{2}F $$ So, apart from the factor of $\tfrac{1}{2}$, the equation your teacher is using works for an object accelerating from rest. That factor of a half is a ...


2

For a particle in a gravitational field treated as a constant? Surely Newton's equations of motion in the fixed rectangular frame: $$\ddot{x}=0$$ $$\ddot{y}=-g$$ are as simple as it can get!


2

Yes, you are thinking about it correctly. No force is required to keep the puck in motion. This is an important idea in physics. It is actually a common misconception among physics students that a force is required to keep an object in motion, so it is good you do not have this misconception.


1

Short Answer: The contact force (normal force if you like) between the pan and the box is 0 because the pan has negligible mass. Long Answer: The key point in this problem is that the pan has neglible mass. Suppose for a second that the pan had some mass $m_p$. After falling a distance of $0.5 \, m$ the box would have velocity $v_b = \sqrt{2gh}$. In the ...


1

With a strong grasp of Lie Algebra and Calculus of variations, "Invariante Variationsprobleme" should provide all the foundation one needs to build Newtonian Mechanics (and so much more). The deeper reason that we use either of these formalism is that they agree with experiment; that either formalism predicts the other is far less valuable than that they ...


1

This type of question keeps reccuring. I suppose either teachers don't teach this correctly, or students do not pay attention in class. $F=m a$ is really $$\sum F = m a_{cm}$$ These two distinctions (the sum, and the acceleration of the center of mass) make all the difference in the world (At least read on Newton's Laws of motion). You push on the rock and ...


1

It is still accelerating, albeit with a lesser magnitude. The derivative of acceleration is called Jerk (See Physical intuition for higher order derivatives). So at point B it is accelerating with negative jerk.


1

However it makes sense that gravity can't travel faster than light because of the force-carrying photons Whilst it makes sense that gravity can't travel faster than light, we don't actually know this for sure. What we do however know is that the force of gravity is not conveyed by photons. Even electromagnetic force is not conveyed by photons - hydrogen ...


1

Just a small conceptual hint will do No problem. A hint: Set up Newton's law, $\sum F=ma$. You will see that the sum of all the three forces must equal... yes, what should it equal? I'm confused with the condition at which the block will start sliding What is the difference in the equation mentioned above for a point just after it started moving, ...


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Assume this is on Earth, so you know $g$. Don't try to use any values for $M$ or $k$. Just use symbols. Draw a free-body diagram for the system at rest. From this you can get the relationship between $M$ and $k$ and $g$. You probably also have an expression telling you the relationship between the $T$, $M$, and $k$. Do the algebra and everything should work ...


1

The ping pong ball would lose a tiny amount of kinetic energy to the truck. The truck ends up with a momentum of just under twice what the ping pong ball had. However, energy is 1/2 m*v^2 = 1/2(m*v)^2/m. Since the truck is much more massive than the ping pong ball, it carries much less energy for a given momentum. The end result is that the small amount of ...


1

Balance torques around the corner of the step, so r x cos(38.7) x mg= F x 0.3. F = 114 N


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Force = mass * acceleration is the basic simplified version of the equation. There are more complected formulas available for this that take into account more complicated scenarios; like taking in the account of different forces coming from different directions like in your scenario. sum(forces) = mass * acceleration force = the derivative of its linear ...


1

The main thing is that the total force you are applying on the body is not enough to move the rock.This means that the total force on the box is zero because the force of static friction is grater than that of your applied force,which cancels out the effect your force.As your forces increases the static friction also increases until a point comes when you ...



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