# 4.7 Further applications of newton’s laws of motion  (Page 4/6)

 Page 4 / 6

So, the scale reading in the elevator is greater than his 735-N (165 lb) weight. This means that the scale is pushing up on the person with a force greater than his weight, as it must in order to accelerate him upward. Clearly, the greater the acceleration of the elevator, the greater the scale reading, consistent with what you feel in rapidly accelerating versus slowly accelerating elevators.

Solution for (b)

Now, what happens when the elevator reaches a constant upward velocity? Will the scale still read more than his weight? For any constant velocity—up, down, or stationary—acceleration is zero because $a=\frac{\Delta v}{\Delta t}$ , and $\Delta v=0$ .

Thus,

${F}_{\text{s}}=\text{ma}+\text{mg}=0+\text{mg}.$

Now

${F}_{\text{s}}=\left(\text{75}\text{.}\text{0 kg}\right)\left(9\text{.}{\text{80 m/s}}^{2}\right),$

which gives

${F}_{\text{s}}=7\text{35 N}.$

Discussion for (b)

The scale reading is 735 N, which equals the person’s weight. This will be the case whenever the elevator has a constant velocity—moving up, moving down, or stationary.

The solution to the previous example also applies to an elevator accelerating downward, as mentioned. When an elevator accelerates downward, $a$ is negative, and the scale reading is less than the weight of the person, until a constant downward velocity is reached, at which time the scale reading again becomes equal to the person’s weight. If the elevator is in free-fall and accelerating downward at $g$ , then the scale reading will be zero and the person will appear to be weightless.

## Integrating concepts: newton’s laws of motion and kinematics

Physics is most interesting and most powerful when applied to general situations that involve more than a narrow set of physical principles. Newton’s laws of motion can also be integrated with other concepts that have been discussed previously in this text to solve problems of motion. For example, forces produce accelerations, a topic of kinematics, and hence the relevance of earlier chapters. When approaching problems that involve various types of forces, acceleration, velocity, and/or position, use the following steps to approach the problem:

Problem-Solving Strategy

Step 1. Identify which physical principles are involved . Listing the givens and the quantities to be calculated will allow you to identify the principles involved.
Step 2. Solve the problem using strategies outlined in the text . If these are available for the specific topic, you should refer to them. You should also refer to the sections of the text that deal with a particular topic. The following worked example illustrates how these strategies are applied to an integrated concept problem.

## What force must a soccer player exert to reach top speed?

A soccer player starts from rest and accelerates forward, reaching a velocity of 8.00 m/s in 2.50 s. (a) What was his average acceleration? (b) What average force did he exert backward on the ground to achieve this acceleration? The player’s mass is 70.0 kg, and air resistance is negligible.

Strategy

1. To solve an integrated concept problem , we must first identify the physical principles involved and identify the chapters in which they are found. Part (a) of this example considers acceleration along a straight line. This is a topic of kinematics . Part (b) deals with force , a topic of dynamics found in this chapter.
2. The following solutions to each part of the example illustrate how the specific problem-solving strategies are applied. These involve identifying knowns and unknowns, checking to see if the answer is reasonable, and so forth.

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Calculate the final velocity attained, when a ball is given a velocity of 2.5m/s, acceleration of 0.67m/s² and reaches its point in 10s. Good luck!!!
2.68m/s
Doc
vf=vi+at vf=2.5+ 0.67*10 vf= 2.5 + 6.7 vf = 9.2
babar
s = vi t +1/2at sq s=58.5 s=v av X t vf= 9.2
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v=u+at where v=final velocity u=initial velocity a=acceleration t=time
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Note: LINEAR MOMENTUM Linear momentum is defined as the product of a system’s mass multiplied by its velocity: size 12{p=mv} {}
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