A boy in a wheelchair (total mass 50.5 kg) has a speed of 1.40 m/s at the crest of a slope 2.40 m high and 12.4 m long. At the bottom of the slope, his speed is 6.50 m/s. Assume air resistance and rolling resistance can be modeled as a constant friction force of 41.0 N. Find the work he did in pushing forward on his wheels during the downhill ride.

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  electron1

  Lv 7

  4 years ago

  There are two forces that cause the wheelchair to accelerate. These two forces are the component of the wheelchair’s weight that is parallel to the hill and the force the boy exerts. The net force on the wheelchair is equal to the sum of these two forces minus the friction force. Let’s use the following equation to determine the acceleration.

   

  vf2 = vi^2 + 2 * a * d

  6.5^2 = 1.4^2 + 2 * a * 12.4

  24.8 * a = 40.29

  a = 40.29 ÷ 24.8

  This is approximately 1.62 m/s^2.

   

  Net force = 50.5 * 40.29 ÷ 24.8 = 2034.645 ÷ 24.8

  This is approximately 82 N.

   

  Force parallel = 50.5 * 9.8 * 2.40 ÷ 12.4 = 1187.76 ÷ 12.4

   

  This is approximately 95.8 N. This force and the force that the boy exerts are the forces that cause the wheelchair to accelerate.

   

  Net force = F + (1187.76 ÷ 12.4) – 41

  Set the two net forces equal to each other.

   

  F + (1187.76 ÷ 12.4) – 41 = 2034.645 ÷ 24.8

  F = 41 + (2034.645 ÷ 24.8) – (1187.76 ÷ 12.4)

  F is approximately 27.3 N

   

  Work = [41 + (2034.645 ÷ 24.8) – (1187.76 ÷ 12.4)] * 12.4 = 337.9625 J

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  JupiterSky

  Lv 6

  4 years ago

  His change in kinetic energy is due to gravitational- and his own work,

  while the friction does negative work.

   

  ∆K = ∆U +∆W + ∆Wf

  ½m(Vf²−Vi²) = mgh + ∆W + Ff*d

  ∆W = ½m(Vf²−Vi²) − mgh − Ff*d

  ∆W = ½*50.5(6.5²−1.4²) − (50.5*9.8*2.4) − (−41*12.4)

  ∆W = 508 J

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asked in Science & MathematicsPhysics · 8 years ago

What is the spring constant? please help me!!!!?

A block of mass 487 g is pushed against the

spring (located on the left-hand side of the

track) and compresses the spring a distance

4.7 cm from its equilibrium position (as shown

in the figure below). The block starts from

rest, is accelerated by the compressed spring,

and slides across a frictionless horizontal track

(as shown in the figure below). It leaves the

track horizontally, flies through the air, and

subsequently strikes the ground.

acceleration = 9.81

 

A) What is the spring constant?

B)What is the speed v of the block when it leaves

the track?

C) What is the total speed of the block when it

hits the ground?

...Show more

Update:

Height Above ground = 1.9 m

Lands = 4.59m away

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What is the spring constant? please help me!!!!?

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  Jánošík

  Lv 7

  8 years ago

  Favorite Answer

  I'm sure there is some crucial information in "the figure below" !!!

   

  Edit:

  OK... now we got something to work with ... ;-)

   

  Time needed to fall a vertical distance of 1.9 m is:

  t = √( 2h / g )

  t = √[ 2×(1.9 m) / (9.81 m/s²) ]

  t = 0.6224 s

   

  In that time, the block must travel a horizontal distance of 4.59 m, so the horizontal velocity is:

  r = v×t

  (4.59 m) = v×(0.6224 s)

  v = 7.37 m/s < - - - - - - - - - - - - - - - - - answer B

   

  That is also the speed of the block as it leaves the spring.

  Its kinetic energy at that time is:

  KE = m×v²/2

  KE = (0.487 kg)×(7.3749 m/s)²/2

  KE = 13.244 J

   

  That is also the elastic potential energy stored in the spring at maximum compression. So the spring constant is:

  KE = PEe = k×d²/2

  (13.244 J) = k×(0.047 m)²/2

  k = 12 N/m < - - - - - - - - - - - - - - - - - - answer A

   

  The gravitational potential energy of the block on the table is:

  PEg = m×g×h

  PEg = (0.487 kg)×(9.81 m/s²)×(1.9)

  PEg = 9.077 J

   

  The total energy of the block on the table is:

  TE = PEg + PEe

  TE = (9.077 J) + (13.244 J)

  TE = 22.32 J

   

  When hitting the ground, all that energy is converted to kinetic energy, so

  TE = KE = m×v²/2

  (22.32 J) = (0.487 kg)×v²/2

  v = 9.57 m/s < - - - - - - - - - - - - - - - - - - answer C

  ...Show more

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  Technobuff

  Lv 7

  8 years ago

  There must be a height above ground and a distance from the end of the track on the "as shown in the figure below".??

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• 

  STALKER

  Lv 4

  8 years ago

  use f=kl

   

  where k is the spring constant and l is extension from the natural length.

   

  k=f/l

   

  since you have mass in grams. use w=mg

   

  w=0.487 x 9.8 = 4.77N

   

  Now use the equation,

   

  k=4.77/4.7

   

  k=1.012

   

  Note: my answer could be wrong since I do not know the springs original length from the diagram shown. But, the method of working out spring constant is same.

  ...Show more

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ひいらぎ

Lv 5

asked in Science & MathematicsPhysics · 1 decade ago

Physics Momentum, Rolling?

Pat builds a track for his model car out of wood. The track is 10.00 cm wide and 3.00 m long, and is solid. The runway is cut such that it forms a parabola by the equation y=(x - 3)^2/9. Locate the horizontal position of the center of gravity of this track.

 

A washing machine load in the "spin" cycle, (much like a centrifuge used to separate fluids), can be modeled as a cylinder which has zero density toward the center, and starting from some inner radius where the clothes (or fluids) begin, increases in density as you move radially outward. The washing machine has density of clothes (in SI units) of p=14r+5, a height of 0.51 m, and only has clothes present from inner radius 0.2m out to 0.7m from the center.

-What is the moment of Inertia of the wash load?

-What is the power of motor that drives the clothes from rest up to 4 rev/sec, in 14 seconds if the basin has moment of Inertia of 4 kg*.m2

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Physics Momentum, Rolling?

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  Anonymous

  1 decade ago

  Favorite Answer

   

  I can't visualize the first problem.

   

  The second: Moment of inertia is I = m r^2. In a shell of thickness dr there is a mass of p * 2 * pi * r * h * dr. The moment of inertia of the shell is 2 pi h p r^3 dr. Now integrate this from r = 0.2m to r = 0.7m.

   

  Kinetic energy is 1/2 m v^2. In rotations the moment of inertia is like the mass and the angular velocity is like the velocity, so rotational energy is 1/2 I omega^2. Omega is in rad/s. The total inertia of the system is the sum of the inertia of the clothes and the basin. Divide the energy by the time needed to get the power.

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asked in Science & MathematicsPhysics · 1 decade ago

Physics:You and your friends push a 75-kg...Help!?

You and your friends push a 75-kg greased pig up an aluminum slide at the county fair, starting from the low end of the slide. The coefficient of kinetic friction between the pig and the slide is 0.070.

 

(a) All of you pushing together (parallel to the incline) manage to accelerate the pig from rest at the constant rate of 4.9 m/s2 over a distance of 1.8 m, at which point you release the pig. The pig continues up the slide, reaching a maximum vertical height above its release point of 50 cm. What is the angle of inclination of the slide? __degree

 

(b) At the maximum height the pig turns around and begins to slip down the slide, how fast is it moving when it arrives at the low end of the slide? __m/s

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Physics:You and your friends push a 75-kg...Help!?

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  Anonymous

  1 decade ago

  Favorite Answer

   

  ♣the distance the pig was pushed up is

  s=0.5*a*t^2, where s=1.8m, a=4.9m/s^2, t is time of push-up;

  speed of the pig gained after s=1.8m is v=a*t; thus v^2=2as;

  ♠ kinetic energy of the pig at the moment it’s released after push-up is

  E=0.5*m*v^2, which is enough for the pig to move by itself L distance more along the ramp, where L=h/sin(b), h=50cm=0.5m, b is angle in question;

  ♦pig’s potential energy at vertex (when stopped) is E1=mgh;

  the energy E2=E-E1 is gone on friction, that is E2=f*L is work of friction, where

  f=μ*w1, w1=mg*cos(b) is component of pig’s weight normal to the aluminum ramp; or;

  f= μ*mg*cos(b);

  ♦Thus E2=E-E1; or; f*L = 0.5*m*v^2 –mgh; or;

  (μ*mg*cos(b)) * (h/sin(b)) =0.5*m* 2as –mgh; or;

  μ*gh*cos(b)/sin(b) =as –gh, hence cot(b) =(as –gh)/(μ*gh), hence

  (a); b=atan(μ*gh/(as-gh))= atan(0.07*9.8*0.5/(4.9*1.8 -9.8*0.5)) =5°;

   

  ♥ pig’s potential energy above the low end of the ramp is

  W= mg*(s*sin(b) +h); pig’s ramp position is p=s +L = s + h/sin(b);

  Pig’s final kinetic energy is W1=W-W2, where W2=f*p is work of friction;

  and W1=0.5*m*v^2; thus

  0.5*m*v^2 = mg*(s*sin(b) +h) – μ*mg*cos(b)*(s + h/sin(b));

  v^2 =2g*(s*sin(b) +h)*(1 –μ*cot(b)) =2.575;

  (b); v =1.605 m/s;

  ...Show more

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Just yesterday I posted a key question as to why the statelottery might hypothetically offer you an upfront lump-sum paymentof $50 million or an ongoing annuity of a $5 million check eachyear for 20 years (totaling $100 million in checks). Remember thatthis was just my simplified hypothetical example. A few peopleattempted the answer to this further above, but here's what'sreally going on...it has to do with Net Present Value (NPV). It'sbecause future-year dollars are worth less, and the further you gointo the future, the less those dollars are worth in terms oftoday's dollars. And that is why, when financial analysts look at adecision involving multi-year cash flows, they will use discountfactors to convert all of the future-year dollars to present value(i.e., this year's dollars) and add them up to yield the NPV of allof the amounts to put a final answer in terms of today's value.Where do these discount factors come from? They're derived from thereciprocal of the famous compound interest formula, but you justneed to go and look at two tables in your textbook in Chapter 12,Appendix 12A, Table A (present value of $1 due in n periods) andTable B (present value of an annuity of $1 per period). Class, Icannot stress to you enough how widely-used this concept is, andthe importance of understanding how these discount factors areapplied to solve for net present value (NPV) and a related metric,internal rate of return (IRR). You will likely need to read, andreread, Chapter 12 and closely walk through some textbook exampleswhere each year's cash flows are multiplied by the appropriatediscount factor to convert all of that money into this year'sdollars (present value). We are surrounded by this financialanalysis methodology when faced with a decision on a lotterypayout, a retirement payout (lump sum versus annuity), and many ofour states and cities right now are dealing with issues relating tothe calculation of NPV in assessing whether they currently haveenough money in the appropriate accounts to meet the futureretirement payment obligations they've made to their publicemployees. This is an economic issue that has been on the brink ofexploding for a long time (don't take my word for it, it's beendiscussed in multiple articles in the NY Times, Wall StreetJournal, The Economist), but I'm not going to add all of that intothis discussion here. Let's learn to walk before we try to run.Would you like to see how the lottery payouts I described would bemultiplied by each year's financial discount factor? Just take alook at my attached Excel workbook, and note that there are threetabs: the first assumed a 0% cost of capital (no financialdiscounting done), the second assumed a 5% cost of capital, and thethird sheet inserted a 10% cost of capital. You'll see how thesedifferent interest rates changed the discount factor multipliersand ultimately changed the NPV of each alternative (the value ofthe lump sum or the annuity, all shown in present value). Also,this example just shows money moving one way (i.e., to the state,it's cash out of their pocket...to the lottery winner, it's cashinto their pocket). In many NPV and IRR examples, money will moveboth ways over multiple years...cash flows that are expenses orinvestments are outflows and are shown as negative, while revenuesor other incoming money are inflows and are shown as positive.Please look closely at my attached Excel workbook (all three tabs)and also walk through all of the examples in the chapter showinghow to do NPV and IRR problems. There are also examples in the textof simpler metrics including payback period and accounting rate ofreturn. It's typically very hard stuff for students to master. Now,for those of you who've looked at the chapter examples, maybeyou've figured this out already:

1. What is the key difference between Tables A and B in Chapter12, Appendix 12A, and how and when are each of them used in ananalysis?

2. What do we mean by “Year 0” in a financial analysis? Whatvalue do we use for the discount factor or multiplier for the cashflow in Year 0 of a financial analysis?

1. You are given only three quarterly seasonal indices and quarterly seasonally adjusted data for the entire year. What is the raw data value for Q4? Raw data is not adjusted for seasonality.

Quarter Seasonal Index Seasonally Adjusted Data

Q1 .80 295

Q2 .85 299

Q3 1.15 270

Q4 --- 271

2. One model of exponential smoothing will provide almost the same forecast as a liner trend method. What are linear trend intercept and slope counterparts for exponential smoothing?

A. Alpha and Delta

B. Delta and Gamma

C. Alpha and Gamma

D. Standard Deviation and Mean

3. When performing correlation analysis what is the null hypothesis? What measure in Minitab is used to test it and to be 95% confident in the significance of correlation coefficient.

A. Ho: r = .05 p < .5

B. Ho: r = 0 p >.05

C. Ho: r ? 0 p?.05

D. Ho: r = 0 p?.05

A. The estimated value is 80% of the average monthly seasonal estimate.
B. The estimate is .80 of the forecasted Y trend value.
C. The estimated value is .80 of the historical average CMA values.
D. The estimated value has 20% more variation than the average historical Y data values.

A. H1: u ? $1,368,000 A one-tailed t-test to the left.
B. H1: u = $1,368,000 A two-tailed t-test.
C. H1: u < $1,368,000 A one-tailed t-test to the left.
D. H1: p < $1,368,000 A one-tailed test to the right