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Calculating distance traveled: how far a baseball player slides

Consider the situation shown in [link] , where a baseball player slides to a stop on level ground. Using energy considerations, calculate the distance the 65.0-kg baseball player slides, given that his initial speed is 6.00 m/s and the force of friction against him is a constant 450 N.

A baseball player slides to stop in a distance d. the displacement d is shown by a vector towards the left and frictional force f on the player is shown by a small vector pointing towards the right equal to four hundred and fifty newtons. K E is equal to half m v squared, which is equal to f times d.
The baseball player slides to a stop in a distance d size 12{d} {} . In the process, friction removes the player’s kinetic energy by doing an amount of work fd size 12{ ital "fd"} {} equal to the initial kinetic energy.

Strategy

Friction stops the player by converting his kinetic energy into other forms, including thermal energy. In terms of the work-energy theorem, the work done by friction, which is negative, is added to the initial kinetic energy to reduce it to zero. The work done by friction is negative, because f size 12{f} {} is in the opposite direction of the motion (that is, θ = 180º size 12{q="180"°} {} , and so cos θ = 1 size 12{"cos"θ= - 1} {} ). Thus W nc = fd size 12{W rSub { size 8{"nc"} } = - ital "fd"} {} . The equation simplifies to

1 2 mv i 2 fd = 0 size 12{ { {1} over {2} }  ital "mv" rSub { size 8{i} rSup { size 8{2} } } - ital "fd"=0} {}

or

fd = 1 2 mv i 2 . size 12{ ital "fd"= { {1} over {2} }  ital "mv" rSub { size 8{i} rSup { size 8{2} } } "." } {}

This equation can now be solved for the distance d size 12{d} {} .

Solution

Solving the previous equation for d size 12{d} {} and substituting known values yields

d = mv i 2 2 f = ( 65.0 kg ) ( 6 . 00 m/s ) 2 ( 2 ) ( 450 N ) = 2.60 m. alignl { stack { size 12{d= { { ital "mv" rSub { size 8{i} rSup { size 8{2} } } } over {2f} } } {} #= { { \( "65" "." 0" kg" \) \( 6 "." "00"" m/s" \) rSup { size 8{2} } } over { \( 2 \) \( "450"" N" \) } } {} # " "=" 2" "." "60 m" "." {}} } {}

Discussion

The most important point of this example is that the amount of nonconservative work equals the change in mechanical energy. For example, you must work harder to stop a truck, with its large mechanical energy, than to stop a mosquito.

Calculating distance traveled: sliding up an incline

Suppose that the player from [link] is running up a hill having a 5 . 00º size 12{5 "." "00"°} {} incline upward with a surface similar to that in the baseball stadium. The player slides with the same initial speed. Determine how far he slides.

A baseball player slides on an inclined slope represented by a right triangle. The angle of the slope is represented by the angle between the base and the hypotenuse, which is equal to five degrees, and the height h of the perpendicular side of the triangle is equal to d sin 5 degrees. The length of the hypotenuse is d.
The same baseball player slides to a stop on a 5.00º size 12{5°} {} slope.

Strategy

In this case, the work done by the nonconservative friction force on the player reduces the mechanical energy he has from his kinetic energy at zero height, to the final mechanical energy he has by moving through distance d size 12{d} {} to reach height h size 12{h} {} along the hill, with h = d sin 5.00º size 12{h=d"sin"5 "." "00"°} {} . This is expressed by the equation

KE i + PE i + W nc = KE f + PE f . size 12{"KE""" lSub { size 8{i} } +"PE" rSub { size 8{i} } +W rSub { size 8{"nc"} } ="KE""" lSub { size 8{f} } +"PE" rSub { size 8{f} } } {}

Solution

The work done by friction is again W nc = fd size 12{W rSub { size 8{"nc"} } = - ital "fd"} {} ; initially the potential energy is PE i = mg 0 = 0 and the kinetic energy is KE i = 1 2 mv i 2 ; the final energy contributions are KE f = 0 size 12{"KE" rSub { size 8{f} } =0} {} for the kinetic energy and PE f = mgh = mgd sin θ size 12{"PE" rSub { size 8{f} } = ital "mgh"= ital "mgd""sin"θ} {} for the potential energy.

Substituting these values gives

1 2 mv i 2 + 0 + ( fd ) = 0 + mgd sin θ.

Solve this for d to obtain

d = 1 2 mv i 2 f + mg sin θ = (0.5) ( 65.0 kg ) ( 6.00 m/s ) 2 450 N + ( 65.0 kg ) ( 9.80 m/s 2 ) sin (5.00º) = 2.31 m. alignl { stack { size 12{d = { { left ( { {1} over {2} } right ) ital "mv""" lSub { size 8{i} } "" lSup { size 8{2} } } over {f+ ital "mg""sin5" rSup { size 8{ circ } } } } } {} #" "= { {0 "." "5 " \( "65" "." "0 kg" \) \( 6 "." "00 m/s" \) rSup { size 8{2} } } over {"450""N "+ \( "65" "." "0 kg" \) \( 9 "." "80 m/s" rSup { size 8{2} } \) "sin 5" rSup { size 8{ circ } } } } {} # " "=2 "." "31 m" "." {}} } {}

Discussion

As might have been expected, the player slides a shorter distance by sliding uphill. Note that the problem could also have been solved in terms of the forces directly and the work energy theorem, instead of using the potential energy. This method would have required combining the normal force and force of gravity vectors, which no longer cancel each other because they point in different directions, and friction, to find the net force. You could then use the net force and the net work to find the distance d size 12{d} {} that reduces the kinetic energy to zero. By applying conservation of energy and using the potential energy instead, we need only consider the gravitational potential energy mgh size 12{ ital "mgh"} {} , without combining and resolving force vectors. This simplifies the solution considerably.

Making connections: take-home investigation—determining friction from the stopping distance

This experiment involves the conversion of gravitational potential energy into thermal energy. Use the ruler, book, and marble from [link] . In addition, you will need a foam cup with a small hole in the side, as shown in [link] . From the 10-cm position on the ruler, let the marble roll into the cup positioned at the bottom of the ruler. Measure the distance d size 12{d} {} the cup moves before stopping. What forces caused it to stop? What happened to the kinetic energy of the marble at the bottom of the ruler? Next, place the marble at the 20-cm and the 30-cm positions and again measure the distance the cup moves after the marble enters it. Plot the distance the cup moves versus the initial marble position on the ruler. Is this relationship linear?

With some simple assumptions, you can use these data to find the coefficient of kinetic friction μ k size 12{μ rSub { size 8{k} } } {} of the cup on the table. The force of friction f on the cup is μ k N size 12{μ rSub { size 8{k} } N} {} , where the normal force N is just the weight of the cup plus the marble. The normal force and force of gravity do no work because they are perpendicular to the displacement of the cup, which moves horizontally. The work done by friction is fd size 12{ ital "fd"} {} . You will need the mass of the marble as well to calculate its initial kinetic energy.

It is interesting to do the above experiment also with a steel marble (or ball bearing). Releasing it from the same positions on the ruler as you did with the glass marble, is the velocity of this steel marble the same as the velocity of the marble at the bottom of the ruler? Is the distance the cup moves proportional to the mass of the steel and glass marbles?

A marble is rolling down a makeshift ramp consisting of a small wooden ruler propped up on one end at about a thirty degree angle. At the bottom of the ramp is a foam drinking cup standing upside-down on its lip. A hole is cut out on one side of the cup so that the marble will roll through the hole when it reaches the bottom of the ramp.
Rolling a marble down a ruler into a foam cup.

Phet explorations: the ramp

Explore forces, energy and work as you push household objects up and down a ramp. Lower and raise the ramp to see how the angle of inclination affects the parallel forces acting on the file cabinet. Graphs show forces, energy and work.

The Ramp

Test prep for ap courses

You are in a room in a basement with a smooth concrete floor (friction force equals 40 N) and a nice rug (friction force equals 55 N) that is 3 m by 4 m. However, you have to push a very heavy box from one corner of the rug to the opposite corner of the rug. Will you do more work against friction going around the floor or across the rug, and how much extra?

  1. Across the rug is 275 J extra
  2. Around the floor is 5 J extra
  3. Across the rug is 5 J extra
  4. Around the floor is 280 J extra

(b)

In the Appalachians, along the interstate, there are ramps of loose gravel for semis that have had their brakes fail to drive into to stop. Design an experiment to measure how effective this would be.

Section summary

  • A nonconservative force is one for which work depends on the path.
  • Friction is an example of a nonconservative force that changes mechanical energy into thermal energy.
  • Work W nc size 12{W rSub { size 8{"nc"} } } {} done by a nonconservative force changes the mechanical energy of a system. In equation form, W nc = Δ KE + Δ PE size 12{W rSub { size 8{"nc"} } =Δ"KE"+Δ"PE"} {} or, equivalently, KE i + PE i + W nc = KE f + PE f size 12{"KE" rSub { size 8{i} } +"PE" rSub { size 8{i} } +W rSub { size 8{"nc"} } ="KE" rSub { size 8{f} } +"PE" rSub { size 8{f} } } {} .
  • When both conservative and nonconservative forces act, energy conservation can be applied and used to calculate motion in terms of the known potential energies of the conservative forces and the work done by nonconservative forces, instead of finding the net work from the net force, or having to directly apply Newton’s laws.

Problems&Exercises

A 60.0-kg skier with an initial speed of 12.0 m/s coasts up a 2.50-m-high rise as shown in [link] . Find her final speed at the top, given that the coefficient of friction between her skis and the snow is 0.0800. (Hint: Find the distance traveled up the incline assuming a straight-line path as shown in the figure.)

A skier is about to go up an inclined slope with some initial speed v sub i shown by an arrow towards right. The slope makes a thirty-five-degree with the horizontal. The height of the point where the slope ends from the skiers’ starting position is two point five meters. Final speed of the skier at the end of the inclined slope is unknown.
The skier’s initial kinetic energy is partially used in coasting to the top of a rise.

9.46 m/s

(a) How high a hill can a car coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110 km/h? (b) If, in actuality, a 750-kg car with an initial speed of 110 km/h is observed to coast up a hill to a height 22.0 m above its starting point, how much thermal energy was generated by friction? (c) What is the average force of friction if the hill has a slope 2 . size 12{2 "." 5°} {} above the horizontal?

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Source:  OpenStax, Work and energy. OpenStax CNX. Nov 09, 2015 Download for free at http://legacy.cnx.org/content/col11902/1.1
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