6.1 Solving problems with newton’s laws  (Page 10/12)

Recall that $v=\frac{ds}{dt}$ and $a=\frac{dv}{dt}$ . If acceleration is a function of time, we can use the calculus forms developed in Motion Along a Straight Line , as shown in this example. However, sometimes acceleration is a function of displacement. In this case, we can derive an important result from these calculus relations. Solving for dt in each, we have $dt=\frac{ds}{v}$ and $dt=\frac{dv}{a}.$ Now, equating these expressions, we have $\frac{ds}{v}=\frac{dv}{a}.$ We can rearrange this to obtain $a^{}ds=v^{}dv.$

Motion of a projectile fired vertically

A 10.0-kg mortar shell is fired vertically upward from the ground, with an initial velocity of 50.0 m/s (see [link] ). Determine the maximum height it will travel if atmospheric resistance is measured as $F_{\text{D}}=(0.0100v^2)\text{N,}$ where v is the speed at any instant.

Strategy

The known force on the mortar shell can be related to its acceleration using the equations of motion. Kinematics can then be used to relate the mortar shell’s acceleration to its position.

Solution

Initially, $y_0=0$ and $v_0=50.0\text{m/s}\text{.}$ At the maximum height $y=h,v=0.$ The free-body diagram shows $F_{\text{D}}$ to act downward, because it slows the upward motion of the mortar shell. Thus, we can write

${\displaystyle \sum }{F}_y=m{a}_y$

$\begin{array}{}\\ \hfill -F_{\text{D}}-w& =\hfill & m{a}_y\hfill \\ \hfill -0.0100v^2-98.0& =\hfill & 10.0a\hfill \\ \hfill a& =\hfill & \mathrm{-0.00100}{v}^2-9.80.\hfill \end{array}$

The acceleration depends on v and is therefore variable. Since $a=f(v)\text{,}$ we can relate a to v using the rearrangement described above,

$a^{}ds=v^{}dv.$

We replace ds with dy because we are dealing with the vertical direction,

$ady=vdv,(\mathrm{-0.00100}{v}^2-9.80)dy=vdv.$

We now separate the variables ( v ’s and dv ’s on one side; dy on the other):

$\begin{array}{}\\ {\displaystyle {\int }_0^{h}dy=}{\displaystyle {\int }_{50.0}^0\frac{vdv}{(\mathrm{-0.00100}{v}^2-9.80)}}\hfill \\ {\displaystyle {\int }_0^{h}dy=}-{\displaystyle {\int }_{50.0}^0\frac{vdv}{(0.00100v^2+9.80)}}=(\mathrm{-5}\times {10}^3)\text{ln}(0.00100v^2+9.80){|}_{50.0}^0.\hfill \end{array}$

Thus, $h=114\text{m}\text{.}$

Significance

Notice the need to apply calculus since the force is not constant, which also means that acceleration is not constant. To make matters worse, the force depends on v (not t ), and so we must use the trick explained prior to the example. The answer for the height indicates a lower elevation if there were air resistance. We will deal with the effects of air resistance and other drag forces in greater detail in Drag Force and Terminal Speed .

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Summary

• Newton’s laws of motion can be applied in numerous situations to solve motion problems.
• Some problems contain multiple force vectors acting in different directions on an object. Be sure to draw diagrams, resolve all force vectors into horizontal and vertical components, and draw a free-body diagram. Always analyze the direction in which an object accelerates so that you can determine whether $F_{\text{net}}=ma$ or $F_{\text{net}}=0.$
• The normal force on an object is not always equal in magnitude to the weight of the object. If an object is accelerating vertically, the normal force is less than or greater than the weight of the object. Also, if the object is on an inclined plane, the normal force is always less than the full weight of the object.
• Some problems contain several physical quantities, such as forces, acceleration, velocity, or position. You can apply concepts from kinematics and dynamics to solve these problems.