Question

3. d) A mass m (in kgs) acted on by a constant force p Newtons, moves a distance x in t sec.
and acquires a velocity v m/ s . Show that
x = mv^2 / 2gp = gt^2p / 2m
where g is the acceleration due to gravity.

Answer 1

Answer on Question #50896 – Math – Differential Calculus | Equations

Question

A mass $m$ (in $kg$) acted on by a constant force $p$. Newtons, moves a distance $x$ in $t$ sec. and acquires a velocity $v_{r\frac{m}{s}}$.

Show that:

$$x = \frac{m v^{2}}{2 g p} = \frac{g t^{2} p}{2 m},$$

where $g$ is the acceleration due to gravity.

Solution

Assumptions:

1. At the initial moment of time ($t = 0$) velocity of the mass is 0.

2. The mass moves only under mentioned above constant force.

Under such assumptions, we can use formulae:

1. $A = Fl$, where $F$ – force, $l$ – distance, $A$ – associated amount of work.

2. $m \ddot{l} = F$, where $m$ – mass, $\ddot{l}$ – acceleration. (Newton’s Law)

3. $A = \frac{m v^{2}}{2}$ (Conservation Law)

Initial conditions:

1. $v(t = 0) = 0$.

2. $l(t = 0) = C$.

Solve differential equation $m \ddot{l} = F$ (second formula):

$\dot{l} = v = \int \frac{F}{m} dt = \frac{F}{m} t + C_1$, because $F$ and $m$ are constants, $C_1$ is an arbitrary real constant;

$l = \int v dt = \int \left(\frac{F}{m} t + C_1\right) dt = \frac{F}{m} \frac{t^2}{2} + C_1 t + C_2$, where $C_1$ and $C_2$ are arbitrary real constants.

Use initial conditions:

$$v(0) = 0, \text{hence } C_1 = 0;$$ $$l(0) = C, \text{hence } C_2 = C.$$

Thus, $l = \frac{F}{m} \frac{t^2}{2} + C$.

Recall that $x = l(t) - l(0)$:

$$x = \frac{F}{m} \frac{t^{2}}{2} + C - C = \frac{F}{m} \frac{t^{2}}{2}$$

Equate expressions for $A$ in the first and the third formula:

$$F l = \frac{m v^{2}}{2},$$

which yields

$$l = \frac{m v^{2}}{2 F},$$

where $l$ is such that $l(t) - l(0) = x$.

Put values of the quantities into obtained formulae:

$$x = \frac{p t^{2}}{2 m}$$ $$x = \frac{m v^{2}}{2 p}$$

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