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Problem exercises

A 5 . 00 × 10 5 -kg size 12{5 "." "00" times "10" rSup { size 8{5} } "- kg"} {} rocket is accelerating straight up. Its engines produce 1 . 250 × 10 7 N size 12{1 "." "250" times "10" rSup { size 8{7} } " N"} {} of thrust, and air resistance is 4 . 50 × 10 6 N size 12{4 "." "50" times "10" rSup { size 8{6} } " N"} {} . What is the rocket’s acceleration? Explicitly show how you follow the steps in the Problem-Solving Strategy for Newton’s laws of motion.

An object of mass m is shown. Three forces acting on it are tension T, shown by an arrow acting vertically upward, and friction f and gravity m g, shown by two arrows acting vertically downward.

Using the free-body diagram:

F net = T f m g = ma size 12{F rSub { size 8{"net"} } =T - f= ital "ma"} {} ,

so that

a = T f mg m = 1 . 250 × 10 7 N 4.50 × 10 6 N ( 5.00 × 10 5 kg ) ( 9. 80 m/s 2 ) 5.00 × 10 5 kg = 6.20 m/s 2 size 12{a= { {T` - `f` - ` ital "mg"} over {m} } = { {1 "." "250" times "10" rSup { size 8{7} } " N" - 4 "." "50" times "10" rSup { size 8{"6 "} } N - \( 5 "." "00" times "10" rSup { size 8{5} } " kg" \) \( 9 "." "80 m/s" rSup { size 8{2} } \) } over {5 "." "00" times "10" rSup { size 8{5} } " kg"} } ="6" "." 20" m/s" rSup { size 8{2} } } {} .

The wheels of a midsize car exert a force of 2100 N backward on the road to accelerate the car in the forward direction. If the force of friction including air resistance is 250 N and the acceleration of the car is 1 . 80 m/s 2 size 12{1 "." "80 m/s" rSup { size 8{2} } } {} , what is the mass of the car plus its occupants? Explicitly show how you follow the steps in the Problem-Solving Strategy for Newton’s laws of motion. For this situation, draw a free-body diagram and write the net force equation.

Calculate the force a 70.0-kg high jumper must exert on the ground to produce an upward acceleration 4.00 times the acceleration due to gravity. Explicitly show how you follow the steps in the Problem-Solving Strategy for Newton’s laws of motion.

  1. Use Newton’s laws of motion.
    Two forces are acting on an object of mass m: F, shown by an arrow pointing upward, and its weight w, shown by an arrow pointing downward. Acceleration a is represented by a vector arrow pointing upward. The figure depicts the forces acting on a high jumper.
  2. Given : a = 4.00 g = ( 4.00 ) ( 9. 80 m/s 2 ) = 39.2 m/s 2 ; size 12{a=4 "." "00" g= \( 4 "." "00" \) \( 9 "." "80 m/s" rSup { size 8{2} } \) ="39" "." 2" m/s" rSup { size 8{2} } " ; "} {} m = 70 . 0 kg size 12{m="70" "." "0 kg"} {} ,

    Find: F size 12{F} .

  3. {} {} F =+ F w = ma , size 12{ Sum {F"=+"F - w= ital "ma"" ,"} } {} so that F = ma + w = ma + mg = m ( a + g ) size 12{F= ital "ma"+w= ital "ma"+ ital "mg"=m \( a+g \) } {} .

    F = ( 70.0 kg ) [ ( 39 . 2 m/s 2 ) + ( 9 . 80 m/s 2 ) ] size 12{F= \( "70" "." 0" kg" \) \[ \( "39" "." "2 m/s" rSup { size 8{2} } \) + \( 9 "." "80 m/s" rSup { size 8{2} } \) \] } {} = 3. 43 × 10 3 N size 12{ {}= {underline {`3 "." "43" times "10" rSup { size 8{3} } " N"}} } {} . The force exerted by the high-jumper is actually down on the ground, but F size 12{F} is up from the ground and makes him jump.

  4. This result is reasonable, since it is quite possible for a person to exert a force of the magnitude of 10 3 N size 12{"10" rSup { size 8{3} } " N"} {} .

When landing after a spectacular somersault, a 40.0-kg gymnast decelerates by pushing straight down on the mat. Calculate the force she must exert if her deceleration is 7.00 times the acceleration due to gravity. Explicitly show how you follow the steps in the Problem-Solving Strategy for Newton’s laws of motion.

A freight train consists of two 8.00 × 10 4 -kg engines and 45 cars with average masses of 5.50 × 10 4 kg . (a) What force must each engine exert backward on the track to accelerate the train at a rate of 5.00 × 10 –2 m/s 2 size 12{5 "." "00" times "10" rSup { size 8{"–2"} } " m/s" rSup { size 8{2} } } {} if the force of friction is 7 . 50 × 10 5 N size 12{7 "." "50" times "10" rSup { size 8{5} } " N"} {} , assuming the engines exert identical forces? This is not a large frictional force for such a massive system. Rolling friction for trains is small, and consequently trains are very energy-efficient transportation systems. (b) What is the force in the coupling between the 37th and 38th cars (this is the force each exerts on the other), assuming all cars have the same mass and that friction is evenly distributed among all of the cars and engines?

(a) 4 . 41 × 10 5 N size 12{4 "." "41" times "10" rSup { size 8{5 } } " N"} {}

(b) 1 . 50 × 10 5 N size 12{1 "." "47" times "10" rSup { size 8{5 } } " N"} {}

Commercial airplanes are sometimes pushed out of the passenger loading area by a tractor. (a) An 1800-kg tractor exerts a force of 1 . 75 × 10 4 N size 12{1 "." "75" times "10" rSup { size 8{4 } } " N"} {} backward on the pavement, and the system experiences forces resisting motion that total 2400 N. If the acceleration is 0 . 150 m/s 2 size 12{0 "." "150 m/s" rSup { size 8{2} } } {} , what is the mass of the airplane? (b) Calculate the force exerted by the tractor on the airplane, assuming 2200 N of the friction is experienced by the airplane. (c) Draw two sketches showing the systems of interest used to solve each part, including the free-body diagrams for each.

Questions & Answers

a perfect square v²+2v+_
Dearan Reply
kkk nice
Abdirahman Reply
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
or infinite solutions?
Kim
y=10×
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if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
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rolling four fair dice and getting an even number an all four dice
ramon Reply
Kristine 2*2*2=8
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Differences Between Laspeyres and Paasche Indices
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No. 7x -4y is simplified from 4x + (3y + 3x) -7y
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is it 3×y ?
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J, combine like terms 7x-4y
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im not good at math so would this help me
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f(x)= 2|x+5| find f(-6)
Prince Reply
f(n)= 2n + 1
Samantha Reply
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
Crystal Reply
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
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preparation of nanomaterial
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Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
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can nanotechnology change the direction of the face of the world
Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
bamidele Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
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Source:  OpenStax, Physics 110 at une. OpenStax CNX. Aug 29, 2013 Download for free at http://legacy.cnx.org/content/col11566/1.1
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