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The object will move in a curved path whose center is not where I am pulling from. This center, my hand and the mass form a triangle whose lead angle might be positive or negative depending if the speed of the mass is increasing or decreasing. Consider the body above at B moving along the indicated curved path (like a closing spiral). While pulling from A ...

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Yes, there are an infinite number of solutions, though your teacher will want you to choose the most obvious one. When the force does work on the mass, that work can be converted into two forms: the potential energy of the object the kinetic energy of the object If you apply a force of $800g$ then once the object has been raised the 2.4m it will still ...

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The time derivative of $v^2$ is $2v \frac{dv}{dt}$ not $2v$. You must use the chain rule.

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Here is the procedure: $KE = 0.5mv^2$ $\frac{d}{dt}KE = 0.5m\frac{d}{dt}v^2$ So the question becomes,how do we find the derivative of $v^2$ with respect to time? One can easily see that $\frac{d}{dt} = \frac{dv}{dt}\frac{d}{dv}$ (Notice how the $dv$ cancels top and bottom) Therefore, $\frac{d}{dt}v^2 = \frac{dv}{dt}\frac{d}{dv}v^2 = \frac{dv}{dt}\times ... 2 A mass is attached to a rope, and put into a circular motion. ... I am applying a force only in the radial direction, so how can the tangential velocity increase if there is no tangential force? For short, I'll call the mass attached to the rope a "rock". So how does the rock gain angular velocity? If you truly are applying a purely radial force, and if ... 2 If$\vec{F}$is a conservative force field, then it satisfies the property $$\tag{1} \vec{\nabla} \times \vec{F} = 0,$$ and it can be written as $$\tag{2} \vec{F} = \vec{\nabla}V,$$ for a scalar function$V$(which corresponds to potential function in physics). Note that, when you put$(2)$into$(1)$it becomes a "curl of a gradient" and is ... 2 "The answer is that since we are proud physicists and not nitpicking mathematicians we will just wing it when the need arises" This quote is taken from A. Zee's Quantum Field Theory in a Nutshell, and it summarizes the attitude of physicists to mathematics. (At least in an undergraduate level) Since we are physicists, most of our mathematics isn't rigorous. ... 1 In mechanics, a mass$m$experiences a force$\textbf{F}$along some path$C$. The work done on the mass is given by $$W = \int_C \textbf{F} \cdot d\textbf{r},$$ such that the energy of the mass increases by$W$. Positive work corresponds to energy being added to the system in question (which is inevitably taken from the surroundings). Edit: To answer ... 1 Only a radial force is applied so that angular momentum$L=rmv$is conserved, $$dL = m(dr\, v+dv\, r)=0.$$ The force imparts an impulse on the system. So the system is starting to move towards a new equilibrium with shorter radius. The constraint $$r\, dv = -v\, dr$$ now means that for negative$dr$the system picks up a positive angular ... 1 When you push the block the block 'pushes' you with the same force and you both gain equal and opposite direction momentums. Both block, and you have now some momentum and hence kinetic energy. The work done on both you and the block is: $$W=\int F \,dx$$ where$F$is a force applied. It must be equal to the total kinetic energy of you and the block (if ... 1 (Wheels will not work on a frictionless surface... But let's assume you mean that surface under the box is frictionless and not the surface under the car) then I will have to keep pressing on the pedal of the car so as to apply a force that continuously counteract Something is wrong here. For equilibrium and a non-moving box, you don't want the box or ... 1 The slope you calculated is$\frac{10 \rm N}{8\rm m}=1.25\,{\rm \frac Nm}$, so the force function becomes$F=1.25\,{\rm \frac Nm}*x$. Hence, the work is $$W=\int{(1.25\,{\rm \frac Nm}\times x)dx}=1.25\,{\rm \frac Nm}\int{xdx}=0.625\,{\rm \frac Nm}x^2$$ (If$F(0)=0$). Because the unit of$x^2$is$\rm m^2$, then the unit of the work is, as expected,$\rm ...

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Is this the correct way to find the derivative of kinetic energy? $$K=\frac{1}{2}m v^2 \\$$ So: $$\frac{dK}{dt} = \frac{1}{2} \left(\frac{dm}{dt} v^2 + 2mv \frac{dv}{dt} \right)$$ If the mass does not change over the time, then $$\frac{dm}{dt}=0$$ And finally $$\frac{dK}{dt} = \frac{1}{2} \left(2mv \frac{dv}{dt} \right)$$ So simplifying:  ...

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