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You are talking about Newton's first law, i.e, inertia (ball should keep moving) This has absolutely nothing to do with the current group-theory tag. The first law doesn't need to explicitly mention friction, it already says "when no force is applied". It's upto you to keep track of external forces. This has been voted for migration to Physics but I doubt ...


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Let us consider there is no external force and the tyres are rolling smoothly without slipping. Here friction doesn't come into play let me explain you how. Even if friction is present, this friction is not the answer. The concept of friction most books provide are deficient. The term "rolling friction" is also a misnomer. The correct answer is rolling ...


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What you need for circular motion is Centripetal Force. Definition: Centripetal force is a force that makes a body follow a curved path: its direction is always orthogonal to the velocity of the body, toward the fixed point of the instantaneous center of curvature of the path. Centripetal force is generally the cause of circular motion. If the road is ...


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You have to split the time domains into the gears needed to reach 60mph. For each gear, there have to be assumptions on the power delivery of the car. Typically 1st gear is traction limited, so you can assume constant acceleration up to the speed where peak power occurs. The relationship between power speed and acceleration is $P(v) = m \,v\, a(v)$. So run ...


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Remember that, if the net force is zero, velocity is constant (not necessarily zero!). You only need to do push a tiny bit harder for a tiny bit of time to start moving the charge. This extra amount can safely be ignored. Once the charge is moving with some nonzero velocity, equal force is enough.


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The answer to this question depends on whether you're working "in real life" or on a physics problem. On a physics problem, the object doesn't start to move. The net force is exactly 0 at this time, so there is no acceleration. In reality, however, the object would possibly move. There's a variety of reasons for this: The normal force, static ...


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The kinetic enegy at $t=0$ is equal to the friction work done when the wheel has stopped: $E_{kin}=W_f$ that is $1/2J \omega_0^2=F_fs=mg \mu 2 \pi r n $ (with $J=mk^2$) where $r$ is the radius of the bearing bore an n the number of revolutions. Solving for the revolutions gives: $n= \frac{J \omega_0^2}{4 \pi m g \mu r}$ The angular acceleration due to ...


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The tyres of the cycle are rolling and the remaining cycle moves with a velocity same as that the centre of mass of the tyres have. Now the question is which force is responsible to bring the cycle at rest. The answer is Air-friction and Rolling-friction. It should be noted that the static and kinetic friction does not come into the picture because the point ...


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Water and air are both weakly magnetic, with magnetic permeabilities on the order of $10^{-6}$ H/m whereas iron, a strongly magnetic material, has a permeability of about 0.25 H/m. Thus, if you could create the supercavitation in water, a magnetic field would do just about nothing to maintain the cavity.


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At low angles when friction can hold the blocks together, then there is no tension on the cable and thus: $$ f_2 = m_A g \sin \theta \\ f_2 = (m_A+m_B) g \sin \theta $$ Only when there is motion on the blocks there is tension. In that case you have $\ddot{x} = \ddot{x}_A = -\ddot{x}_B$ $$ f_2 = \mu_S m_A g \cos \theta \\ f_2 = -\mu_S (m_A+m_B) g \cos ...


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Correction, the fourth equation should be $T+f_1+f_2=m_BgSin\theta$ Tension will be produced in the string if the blocks are stretched from the other side. When $\theta \leq 21.8^\circ$, you calculated that the frictional force on $A$ will be more than the force due to the incline. So yes, $A$ will accelerate upwards, slacking the string, so no Tension will ...


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Hint : Friction opposes tendency to move. Tension is produced if the string is stretched $very$ slightly. So, increase friction to maximum and then tension will act if necessary. Ironically, you are thinking absolutely right. Give yourself a cookie. From part $a$, we know that the blocks will be at rest at all angles below that. You are also right as at ...


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First remember that it isn't the acceleration/force that matters. It's the velocity -- friction opposes motion, not acceleration. So if the block starts with a upward motion, the friction will point down, no matter what the forces are. So you definitely need to know about the direction of the velocity. If you have an initial velocity, this isn't a problem ...


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Since we know that the block is restricted to move only along the inclined plane, all the force components perpendicular to the inclined plane come in action-reaction pairs( assuming that the inclined plane is stationary and perfectly rigid ) and therefore I shall not include those components in my free body diagrams in order to make things clear. First, ...



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