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7/30/2019 Work, Energy & Power_answer
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MARK SCHEME Tutorial 3 Work, Energy & Power
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1. (a) Comment on use of weighing
Clear statement correctly identifying weight or mass (or their units)e.g. kg a unit of mass, not weight (1) 1
(b) Calculation to check statement
Use of equation of motion to show time or distance (1) Answer to 2 sig figs [120 m or 4.5 s] [no ue] (1)
Example of calculation:
s = ut + at 2 s = 0 + 9.81 m s 2 (5s) 2 OR 100 = 0 + 9.81 m s 2 t 2 s = 123 m OR t = 4.5 s 2
(c) Calculation of kinetic energyEither Use of equation(s) of motion which allow(s) v2 or v to be found (1) Recall of ke = mv2 (1) Answer [69 000 J] (1)
OR Recall of E p = mgh (1) Substitution (1) Answer [69 000 J] (1)
Example of calculation: 2 = u2 + 2 as 2 = 0 + 2 9.81 m s 2 100 m 2 = 1962 m 2 s 2 ke = 1/2 mv2 = 69 000 J (68 670 J)
OR gpe = mghgpe lost = 70 kg 9.81 N kg 1 100 mgpe lost = 69 000 J (68 670 J)[so ke = 69 000 J because ke gained = gpe lost] 3
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2. (i) Work done
Use of work done = force distance (1)Answer given to at least 3 sig fig. [2396 J, 2393 J if 9.8 m s 2 is used, (1) 22442 J if g = 10 m s 2 is used. No ue.]
Work done = 110 kg 9.81 m s 2 2.22 m= 2395.6 J
7/30/2019 Work, Energy & Power_answer
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MARK SCHEME Tutorial 3 Work, Energy & Power
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(ii) Power exerted
Use of power =time
donework or power = F v (1)
Answer: [799 W. 800 W if 2400 J is used and 814 W if 2442 J is 2used. Ecf value from (i)] (1)
Power =3s
J2396
= 798.6 W
(iii) Principle of Conservation of Energy
Either Energy can neither be created nor destroyed (1) (1) OREnergy cannot be created/destroyed or total energy is not lost/gained (1) (merely) transformed from one form to another or in a closed/isolatedsystem. (1) 2
[Simple statement Energy is conserved gets no marks][Information that is not contradictory ignore. Q = U + W, withterms defined acceptable for 1st mark]
(iv) How principle applied to...
Lifting the bar: -Chemical energy (in the body of the weightlifter) or work done(lifting bar) = (gain in) g.p.e. (of bar) (1)
[Reference to k.e. is acceptable]The bar falling: -Transfer from g.p.e. to k.e. (1) (and that) g.p.e. lost = k.e. gained (1) 3
[g.p.e. converted to k.e. would get one mark][References to sound and thermal energy are OK, but gpe to sound or thermal energy on its own gets no marks]
(v) Speed of bar on reaching the floor Setting mv2 = m g h or mv2 = work done or 2400 J (1)
[ecf their value][Shown as formulae without substitution or as numbers substitutedinto formulae]Correct values substituted (1) [allow this mark if the 110 kg omitted substitution gives v2 = (1)
43.55(6) m 2 s 2 or 44.4 m 2 s 2 if g = 10 m s 2 is used]Answer: [6.6 m s 1. 6.7 m s 1 if g = 10 m s 2 is used.]
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MARK SCHEME Tutorial 3 Work, Energy & Power
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110 kg v2 = 110 kg 9.81 m s 2 2.22 m or = 2400 J / 2396 Jv = 6.6 m s 1 [6.66 m s 1 if 10 m s 2 used] (1)
ORSelects v2 = u 2 + 2as or selects 2 relevant equations (1)Correct substitution into equation (1) Answer [6.6 m s 1] (1)
v2 = 0. + 2 9.81 ms 2 2.22m 3v = 6.6 m s 1
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3. (a) (i) Explain the shape of the graph in the part labelled AB
Force proportional to extension / obeys Hookes law (1) 1
(ii) Explain what is happening in the part of the graph labelled CD.
Fully compressed / coils closed (accept cup/bug/toy touches base) (1) 1
(b) Show that the stiffness of the spring is about 1000 N m 1 .
State k = 1/gradient or use of values in k = F / x (1) Correct answer to at least 2 s.f. [1100 N m 1] (1) (Values from graph must be within half a square)(Accept 1000 N m 1 to only 1 s.f. if the answer given by the values
used from the graph is 1.0 103
N m 1
to 2 s.f.) 2Example of calculation k = F / x = 20 N / 0.019 m= 1050 N m 1
(c) (i) Calculate the energy stored in the spring at this stage
State area under graph or use of energy = 1/2 F x or state energy =1/2 kx2 (1) correct answer [0.17 J] (1) [ecf for k ](Values from graph must be within half a square) 2
Example of calculation energy = 1/2 F x = 1/2 19.2 N 0.018 m= 0.17 J
(ii) Calculate the maximum height reached by the bug.Use of gpe = mgh (1) correct answer [2.4 m] (1) [ecf]
Example of calculation 0.17 J = mgh h = 0.17 m / 7.3 10 3 kg x 9.81 N kg 1 = 2.4 m 2
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(iii) State an assumption made in your calculation
all elastic pe ke of bug gpe of bug (2 out of 3) /all stored energy (of the spring) transferred to the toy /no energy lost due to air resistance (1) 1
(d) Explain the advantage of using the video cameraimproves accuracy/reliability/precision (1) eliminate reaction time in looking / can slow down and stop (to takereading) etc (1) 2
(e) Comment on this data
Has not included 0.36 / has not included the anomalous result /0.36 is anomalous/outlier etc (1) 1
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4. Show that average daily capacity provides about 2 10 13 J
E p = mgh (1)
= (28 10 6 m3 10 3 kg m 3) 9.81 N kg 1 64.5 m
= 1.8 10 13 J [no up] (1) 2
Calculation of efficiency over one year
Efficiency = (useful energy output/total energy input) 100%6.1 10 15 J (1) 365 1.77 10 13 J (1) 100%
= 94.4 % [Accept fractional answers. Allow use of 2 10 13 J, whichgives 83.6%, or ecf, but check nos.] (1) 3
Calculation of average power output over year
P = W/t (1)
= 6.1 10 15 J 3.16 10 7 s
= 1.9 108
W (1) 2Reason for difference from max power output
Any sensible reason, e.g., river flow varies over the year / variationsin rainfall [Accept answers related to demand] (1) 1
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MARK SCHEME Tutorial 3 Work, Energy & Power
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5. Calculate kinetic energy
E k = m 2 (1)
E k = 1800 kg (53 m s 1)2
= 2.53 10 6 J (1) 2
Show that max height would be about 140 m
E p = mg h (1)
Initial E k = final E p/ m 2 = mg h/2.53 10 6 J = mg h (1)
h = 2.53 10 6 J/(1800 kg 9.81 N kg 1)h = 143 m [no ue] (1)
OR
2
= u2
+ 2 as0 m 2 s 2 = (53 m s 1)2 + 2 ( 9.81 m s 2) s [subst] (1)
s = (53 m s 1)2 (2 9.81 m s 2) [rearrangement] (1)
s = 143 m [no ue] (1) 3
Show that energy loss is about 3 10 5 J
E p = 1800 kg 9.81 N kg 1 126 m = 2.22 10 6 J (1)
E k E p = 2.53 106 J 2.22 10 6 J
= 3.1 10 5 J [no ue] (1)
OR
For 143 m
E p = 1800 kg 9.81 N kg 1 143 m = 2.53 10 6 J
For 126 m
E p = 1800 kg 9.81 N kg 1 126 m = 2.22 10 6 J (1)
Energy lost =2.53 10 6 J 2.22 10 6 J
= 3.1 10 5 J [no ue] (1)
OR
Energy lost = 1800 kg 9.81 N kg 1 (143 m 126 m) (1)= 3.1 10 5 J [no ue] (1) 2
Calculation of average resistive force
Work = force distance (1)
Force = work distance
= 3.1 10 5 J 126 m
= 2500 N (1) 2
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Calculation of time for climb
s = ( u + v) t (1)
t = 2 s (u + v)
= 2 126 m 53 m s 1
= 4.8 s (1)[Use of g = 9.81 m s 2 in equations of motion to get a consistent valueof t [ = u + at t = 5.4 s] 1 mark]
Assumption: eg assume uniform acceleration/constant resistive force/constant frictional force (1) 3
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6. Show that expected speed is about 35 m s 1
Ek = mv2 and E p = mg h (1)
mv2 = mg h (1)
v = gh2
= ).( m64kg N8192 1
= 35.4 ms 1 [No ue] (1) 3
[For v2 = u 2 + 2as mark u = 0 (1) , rest of substitution 1) , evaluation (1) ]
Assumption
No resistive force, all gpe ke, constant acc n (1) 1
[Do not accept g = 9.81 m s 2]
Reason for lower speed
Work done against resistive force/frictional forces oppose motion/ (1) some g.p.e. heat/sound ...
reduces maximum kinetic energy / acceleration is reduced/less than (1) 2 9.8 m s 2
Calculate efficiency
Efficiency = (actual max k.e. theoretical max k.e.) 100%OR efficiency = (actual max k.e. initial p.e.) 100% (1)
= (m act2 m th
2) 100% OR = ( m act2) (mgh) 100%
21
2 1
)sm435(
)sm532(
.
. 100% =
648.9)sm532( 2 1 .
100% (1)
= 84.2% (1) 3
Reason why speed greater than expected
e.g. motor assisted / initial speed > 0 / run up before drop (1) 1[10]
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7. (a) Sketch a vector diagram
Correct diagram closed polygon, accept a triangle using theresultant of lift and weight, but arrows must follow correctly.Must show sequence of tip-to-tail arrowed vectors. 1
(b) Find the tension in the string.Use of trigonometrical function for the horizontal angle (allowmark for vertical angle if correct and shown on dia) 1
Correct answer for horizontal angle (32.8) 1
Use of Pythagoras or trigonometrical function for the tension 1
Correct answer for tension magnitude (7.1 N) 1
Example of calculation weight lift = 3.86 Nfrom horizontal, tan (angle) = 3.86 N/ 6.0 Nangle = 32.8T 2 = F h
2 + F v2
= (6.0 N) 2 + (3.86 N) 2 T = 7.1 N
(c) (i) Calculate the work done by the girl.
Use of W = Fs 1
Correct answer (150 J) 1
Example of calculation W = Fs = 6.0 N 25 m= 150 J
(ii) Calculate rate at which work is done
Finds time 1
Correct rate (12 W) 1
Example of calculation t = s/v = 25 m / 2.0 m s 1 = 12.5 s
P = 150 J / 12.5 s= 12 W
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8. (a) Explain this demonstration and the need for the precautions.
QWC spelling of technical terms must be correct and theanswer must be organised in a logical sequence
Max 4 from this partIt will not strike the students face / at most will just touch /
returns to starting point (1) The total energy of the pendulum is constant / energy isconserved (1) It cannot move higher than its starting point (1) because that would require extra gpe (consequent on
previous mark) (1) Mention specific energy transfer: gpe ke / ke gpe (1) Energy dissipated against air resistance (1) will stop it quite reaching its starting point (consequent onattempt at describing energy loss mechanism) (1)
Max 4 from this partPushing does work on the ball / pushing provides extra energy (1) If pushed, it can move higher (accept further) (1) will hit the student (1) If the face moves (forward) the ball may reach it (before it isat its maximum height) OR if the face moves (back) the ballwont reach it (1) Max 6
(b) (i) Calculate the gravitational potential energy gained by the ball.
Use of gpe = mgh 1
Correct answer (100 J) 1
Example of calculation gpe = mgh = 7 kg 9.81 N kg 1 1.5 m= 103 J
(ii) Calculate the speed of the ball at the bottom of its swing
Use of ke = 1/2 mv 2 1
Correct answer (5.4 m s 1) 1
Example of calculation 103 J = 1/2 mv2 v = (2 103 J / 7 kg)= 5.4 m s 1 (Use of 100 J 5.3 m s 1)
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