Q&A

# Post History

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posted 10mo ago by Derek Elkins‭  ·  edited 10mo ago by Derek Elkins‭

#2: Post edited by Derek Elkins‭ · 2021-12-14T11:33:21Z (10 months ago)
• Your analysis seems pretty good. To take an equivalent but more traditional example, imagine we throw a ball upwards. Ignoring air resistance and approximating the gravitational force as constant, the ball will accelerate downward with constant acceleration. If the initial (upwards) velocity is $v_0$ and the downward acceleration is $a=-g$, the ball will reach the apex at time $t = v_0/a$ at which point it will have zero velocity and thus zero kinetic energy. It will have traveled a distance of $v_0^2/(2a)$. In general, work is force applied over a distance. In this case, the work done by the gravitational field is the (constant) force, $F = -mg$, applied over this distance, i.e. $mv_0^2/2$.
• Moving more towards your analysis, once we fix a constant force (and constant mass), then we know work is directly proportional to the distance the mass travels, i.e. $W=Fd$. Your analysis shows that $d$ is quadratic in the initial velocity, thus work is quadratic in the initial velocity. I believe this was your goal. The slightly more detailed analysis in the previous paragraph will reproduce exactly the kinetic energy expression if you wanted more.
• Your analysis seems pretty good. To take an equivalent but more traditional example, imagine we throw a ball upwards. Ignoring air resistance and approximating the gravitational force as constant, the ball will accelerate downward with constant acceleration. If the initial (upwards) velocity is $v_0$ and the downward acceleration is $a=-g$, the ball will reach the apex at time $t = v_0/a$ at which point it will have zero velocity and thus zero kinetic energy. It will have traveled a distance of $v_0^2/(2a)$. In general, work is force applied over a distance. In this case, the work done by the gravitational field is the (constant) force, $F = -mg$, applied over this distance, $v_0^2/(2a)$, giving a work done of $mv_0^2/2$.
• Moving more towards your analysis, once we fix a constant force (and constant mass), then we know work is directly proportional to the distance, $d$, the mass travels, i.e. $W=Fd$. Your analysis shows that $d$ is quadratic in the initial velocity, thus work is quadratic in the initial velocity. I believe this was your goal. The slightly more detailed analysis in the previous paragraph will reproduce exactly the kinetic energy expression if you wanted more.
#1: Initial revision by Derek Elkins‭ · 2021-12-14T11:28:57Z (10 months ago)
Your analysis seems pretty good. To take an equivalent but more traditional example, imagine we throw a ball upwards. Ignoring air resistance and approximating the gravitational force as constant, the ball will accelerate downward with constant acceleration. If the initial (upwards) velocity is $v_0$ and the downward acceleration is $a=-g$, the ball will reach the apex at time $t = v_0/a$ at which point it will have zero velocity and thus zero kinetic energy. It will have traveled a distance of $v_0^2/(2a)$. In general, work is force applied over a distance. In this case, the work done by the gravitational field is the (constant) force, $F = -mg$, applied over this distance, i.e. $mv_0^2/2$.
Moving more towards your analysis, once we fix a constant force (and constant mass), then we know work is directly proportional to the distance the mass travels, i.e. $W=Fd$. Your analysis shows that $d$ is quadratic in the initial velocity, thus work is quadratic in the initial velocity. I believe this was your goal. The slightly more detailed analysis in the previous paragraph will reproduce exactly the kinetic energy expression if you wanted more.