Q&A

# Post History

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

• 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.
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.