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Solution for (a)

We are given the initial and final velocities (zero and 8.00 m/s forward); thus, the change in velocity is Δ v = 8.00 m/s . We are given the elapsed time, and so Δ t = 2.50 s size 12{Δt=2 "." "50"" s"} {} . The unknown is acceleration, which can be found from its definition:

a = Δ v Δ t size 12{a= { {Δv} over {Δt} } } {} .

Substituting the known values yields

a = 8.00 m/s 2 . 50 s = 3 . 20 m/s 2 .

Discussion for (a)

This is an attainable acceleration for an athlete in good condition.

Solution for (b)

Here we are asked to find the average force the player exerts backward to achieve this forward acceleration. Neglecting air resistance, this would be equal in magnitude to the net external force on the player, since this force causes his acceleration. Since we now know the player’s acceleration and are given his mass, we can use Newton’s second law to find the force exerted. That is,

F net = ma size 12{F rSub { size 8{"net"} } = ital "ma"} {} .

Substituting the known values of m size 12{m} {} and a size 12{a} {} gives

F net = ( 70.0 kg ) ( 3 . 20 m/s 2 ) = 224 N .

Discussion for (b)

This is about 50 pounds, a reasonable average force.

This worked example illustrates how to apply problem-solving strategies to situations that include topics from different chapters. The first step is to identify the physical principles involved in the problem. The second step is to solve for the unknown using familiar problem-solving strategies. These strategies are found throughout the text, and many worked examples show how to use them for single topics. You will find these techniques for integrated concept problems useful in applications of physics outside of a physics course, such as in your profession, in other science disciplines, and in everyday life. The following problems will build your skills in the broad application of physical principles.

Summary

  • Newton’s laws of motion can be applied in numerous situations to solve problems of motion.
  • Some problems will 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 net = ma size 12{F rSub { size 8{"net"} } = ital "ma"} {} or F net = 0 size 12{F rSub { size 8{"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, the normal force will be less than or greater than the weight of the object. Also, if the object is on an inclined plane, the normal force will always be less than the full weight of the object.
  • Some problems will contain various physical quantities, such as forces, acceleration, velocity, or position. You can apply concepts from kinematics and dynamics in order to solve these problems of motion.

Conceptual questions

To simulate the apparent weightlessness of space orbit, astronauts are trained in the hold of a cargo aircraft that is accelerating downward at g size 12{g} {} . Why will they appear to be weightless, as measured by standing on a bathroom scale, in this accelerated frame of reference? Is there any difference between their apparent weightlessness in orbit and in the aircraft?

A cartoon shows the toupee coming off the head of an elevator passenger when the elevator rapidly stops during an upward ride. Can this really happen without the person being tied to the floor of the elevator? Explain your answer.

Problem exercises

A 76.0-kg person is being pulled away from a burning building as shown in [link] . Calculate the tension in the two ropes if the person is momentarily motionless. Include a free-body diagram in your solution.

An object of mass m is shown being pulled by two ropes. Tension T sub two acts toward the right at an angle of ten degrees above the horizontal. Another rope makes an angle fifteen degrees to the left of the vertical direction, and tension in the rope is T sub one, shown by a vector arrow. Weight w is acting vertically downward.

T 1 = 736 N size 12{T rSub { size 8{1} } ="736"" N"} {}

T 2 = 194 N size 12{T rSub { size 8{2} } ="194 N"} {}

A lady is being pulled away from a burning building using a rope. She is in the middle of the rope; her weight is shown by a vector acting vertically downward. Tension, T sub one, acts upward through the left side of the rope, making an angle of fifteen degrees with the vertical. Tension T sub two acts through the right side of the rope, making an angle of ten degrees above the positive x axis.
The force T 2 size 12{T rSub { size 8{2} } } {} needed to hold steady the person being rescued from the fire is less than her weight and less than the force T 1 size 12{T rSub { size 8{1} } } {} in the other rope, since the more vertical rope supports a greater part of her weight (a vertical force).

Integrated Concepts When starting a foot race, a 70.0-kg sprinter exerts an average force of 650 N backward on the ground for 0.800 s. (a) What is his final speed? (b) How far does he travel?

(a) 7.43 m/s size 12{7 "." "43"" m/s"} {}

(b) 2.97 m

Integrated Concepts A basketball player jumps straight up for a ball. To do this, he lowers his body 0.300 m and then accelerates through this distance by forcefully straightening his legs. This player leaves the floor with a vertical velocity sufficient to carry him 0.900 m above the floor. (a) Calculate his velocity when he leaves the floor. (b) Calculate his acceleration while he is straightening his legs. He goes from zero to the velocity found in part (a) in a distance of 0.300 m. (c) Calculate the force he exerts on the floor to do this, given that his mass is 110 kg.

(a) 4.20 m/s size 12{4 "." "20"" m/s"} {}

(b) 29.4 m/s 2 size 12{"29" "." 4" m/s" rSup { size 8{2} } } {}

(c) 4 . 31 × 10 3 N size 12{4 "." "31" times "10" rSup { size 8{3} } " N"} {}

Unreasonable Results A 75.0-kg man stands on a bathroom scale in an elevator that accelerates from rest to 30.0 m/s in 2.00 s. (a) Calculate the scale reading in newtons and compare it with his weight. (The scale exerts an upward force on him equal to its reading.)

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Source:  OpenStax, Newton's laws. OpenStax CNX. Oct 25, 2015 Download for free at https://legacy.cnx.org/content/col11898/1.1
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