Title 5. P1C5.-Work-Power-Energy-Engg. (With Solve).pdf ✅

KvR, kw3 I ÿgZv  Engineering Practice Sheet ............................................................................................................. 1 WRITTEN weMZ mv‡j BUET-G Avmv cÖkœvejx 1. gvby‡li ürwcÐ 80 mm cvi` Pv‡ci wecix‡Z cÖwZ ̄ú›`‡b 70 mL i3 wbtmiY K‡i| bvwoi K¤úv1⁄4 cÖwZ wgwb‡U 90 n‡j ürwc‡Ði ÿgZv KZ? (cvi‡`i NbZ¡ 13.6 g/cc) [BUET 23-24 cÖ‡kœi Abyiƒc, 21-22] mgvavb: ÿgZv, P = W t = Pvc  AvqZ‡bi cwieZ©b t = hg  V t = 0.08  13600  9.8  70  10–6 60 90 W = 1.12 W (Ans.) 2. †KvwfW-19 (COVID-19) AwZgvix mg‡q 4  103 kg f‡ii GKwU wcKAvc UavK Ges 103 kg f‡ii GKwU gvB‡μvevm Aw·‡Rb wmwjÛvi mieiv‡ni Rb ̈ GKwU nvmcvZv‡j hvw”Qj| wcKAvc Uav‡Ki †eM NÈvq 80 km wQj| GKB MwZkw3 m¤úbœ n‡Z n‡j gvB‡μvevm‡K KZ †e‡M Pj‡Z n‡e? [BUET 20-21] mgvavb: EkM = EkT  1 2 mMv 2 M = 1 2 mTv 2 T  103  v 2 M = 4  103  (80)2  vM = 160 kmh–1 (Ans.) 3. †Kv‡bv Kzqv †_‡K 30 m Dc‡i cvwb †Zvjvi Rb ̈ 5 kW Gi GKwU cv¤ú e ̈envi Kiv nq| cv‡¤úi Kg©`ÿZv 90% n‡j cÖwZ wgwb‡U KZ wjUvi cvwb †Zvjv hv‡e? [g = 9.8 ms–2 ] [BUET 18-19] mgvavb: Pt = mgh  (5  103  0.9)  60 = m  9.8  30  m = 918.37 g V = m  = 918.37 1 L = 918.37 L (Ans.) 4. 1 wU †μb cÖwZwU 50 kg IR‡bi 12 wU wm‡g‡›Ui e ̈vM mg`aæwZ‡Z 160 m DuPz GKwU wbg©vbvaxb fe‡bi Qv‡` IVv‡Z 1 min 10 sec mgq †bq| †μbwUi ÿgZv Ak¦kw3‡Z †ei Ki| [BUET 17-18] mgvavb: Avgiv Rvwb, ÿgZv, P = Fv GLv‡b, F = nmg = 12  50  9.8 N = 5880 N v = s t = 160 m 70 s = 16 7 ms–1  †μ‡bi ÿgZv, P = Fv = 5880  16 7 W = 13440 W = 13440 746 HP = 18.02 HP (Ans.) 5. 3 kg f‡ii e ̄‘i Dci GKwU ej wμqvkxj Av‡Q| e ̄‘wUi Ae ̄’v‡bi mgxKiY x = 3t – 4t2 + t3 , †hLv‡b x Gi gvb wgUv‡i Ges t Gi gvb †m‡K‡Û| t = 0 n‡Z t = 4 †m‡KÛ mg‡q ejwU Øviv e ̄‘i Dci K...ZKv‡Ri cwigvY wbY©q K‡iv| [BUET 16-17] mgvavb: x = 3t – 4t2 + t3  v = dx dt = 3 – 8t + 3t2 t = 0 s G v0 = 3 ms–1 t = 4 s G v4 = 19 ms–1  W = Ek = 1 2  3  (192 – 3 3 ) = 528 J (Ans.) 6. GKwU BwÄb 200 m Mfxi K‚c †_‡K cÖwZ wgwb‡U 500 kg cvwb D‡Ëvjb K‡i| hw` 20% ÿgZvi AcPq nq Zvn‡j BwÄbwUi cÖK...Z ÿgZv KZ? [BUET 12-13] mgvavb: Pout = Pin    mgh t = Pin    Pin = mgh t  1  = (500  9.8  200) 60  1 80% W = 20416.67 W (Ans.)
2 .........................................................................................................................................  Physics 1st Paper Chapter-5 7. 1200 kg f‡ii GKwU Mvoxi Bwćbi ÿgZv 134.05 HP I Kg©`ÿZv 90%| MvwowU‡K w ̄’ive ̄’v †_‡K 30 ms–1 †e‡M Avb‡Z b~ ̈bZg KZ mgq jvM‡e? (1 HP = 0.746 kW) [BUET 10-11] mgvavb: cÖvß ÿgZv, P = 134.05  746  90% W = 90001.17 W MwZkw3i cwieZ©b = K...ZKvR  1 2 mv2 = Pt  t = 1 2P mv2  t  6 sec (Ans.) 8. GKwU ey‡jU GKwU †`qv‡ji g‡a ̈ 0.06 m cÖ‡ek Kivi ci Gi Avw`‡e‡Mi A‡a©K nvivq| ey‡jUwU †`qv‡ji g‡a ̈ Avi KZ`~i cÖ‡ek Ki‡Z cvi‡e? [BUET 08-09] mgvavb: a = v 2 – v 2 4 2  0.06 = 25 4 v 2 s = v 2 4 – 0 25v2 4  2 = 0.02 m (Ans.) 9. 2 mm e ̈vmv‡a©i GKwU e„wói †duvUv 250 m D”PZv †_‡K gvwUi Dci co‡Q| e„wói †duvUvi Dci AwfKl©xq ej KZUv KvR Ki‡e? [BUET 07-08] mgvavb: W = mgh = 4 3 r 3 gh = 4 3    (2  10–3 ) 3  1000  9.8  250 J = 0.0821 J (Ans.) 10. 8 kg f‡ii GKwU e ̄‘ 10 m Dci n‡Z c‡o evwj‡Z 50 cm cÖ‡ek K‡i †_‡g †Mj| e ̄‘wUi Dci evwji Mo evav wbY©q Ki| [BUET 05-06] mgvavb: mg (h + x) = Fx  F = 8 × 9.8 × (10 + 0.5) 0.5  F = 1646.4 N weMZ mv‡j CKRUET-G Avmv cÖkœvejx 1. Avbyf‚wgK Kv‡Vi Dci GKwU †c‡iK Djø¤^fv‡e ivLv Av‡Q| 1 kg f‡ii GKwU nvZzwo Øviv †c‡iKwU‡K Lvov wb‡Pi w`‡K 4 ms–1 †e‡M AvNvZ Kiv nj| †c‡iKwU Kv‡Vi g‡a ̈ 0.015 m Xz‡K †M‡j Mo evav`vbKvix ej wbY©q K‡iv| [CKRUET 20-21; RUET 05-06] mgvavb: Fh = 1 2 mv2 + mgh  F  0.015 =     1 2  1  4 2 + (1  9.8  0.015)  F = 543.2 N (Ans.) weMZ mv‡j KUET-G Avmv cÖkœvejx 1. GKwU e ̄‘‡K wbw`©ó D”PZv †_‡K †d‡j †`qv nj| f‚wg n‡Z 10 wgUvi D”PZvq MwZkw3 w ̄’wZkw3i wØ ̧Y n‡j KZ D”PZv †_‡K e ̄‘wU‡K †djv n‡qwQj? [KUET 19-20] mgvavb: Ek = 2Ep  mg(h – 10) = 2  mg  10  h – 10 = 20  h = 30 m 2. GKwU cvwbc~Y© cyKz‡ii MfxiZv 10 wgUvi, •`N© ̈ 10 wgUvi Ges cÖ ̄’ 5 wgUvi| 30 wgwb‡U cyKziwU‡K cvwb k~b ̈ Ki‡Z KZ Ak¦ÿgZvi cv¤ú e ̈envi Ki‡Z n‡e? [g = 980 cms–2 ] [KUET 19-20] mgvavb: P = mgh t = Vgh t = 103  (10  10  5)  9.8  10 2 30  60  746 HP = 18.25 HP (Ans.) 3. 90 dzU D”PZv n‡Z GKwU e ̄‘‡K cwZZ n‡Z †`qv nj| †Kv_vq Gi MwZkw3 w ̄’i kw3i A‡a©K n‡e? [KUET 04-05] mgvavb: Ek = 1 2 EP  90 – x = x. 1 2  x = 60 ft (Ans.) 4. 300 m DuPz n‡Z GKwU e ̄‘ AwfK‡l©i Uv‡b gy3fv‡e wb‡P co‡j †Kv_vq Zvi MwZkw3 w ̄’wZkw3i A‡a©K n‡e? [KUET 04-05] mgvavb: Ek = 1 2 Ep  300 – x = 1 2 x  x = 200 m (f‚wg n‡Z) (Ans.)
KvR, kw3 I ÿgZv  Engineering Practice Sheet ............................................................................................................ 3 weMZ mv‡j RUET-G Avmv cÖkœvejx 1. GKwU `vjv‡bi Qv‡`i mv‡_ jvMv‡bv 5 m j¤^v GKwU gB Avbyf‚wg‡Ki mv‡_ 30 †KvY K‡i Av‡Q| 60 kg f‡ii GK e ̈w3 25 kg f‡ii IRbmn 10 sec G Qv‡` DV‡j Zvi Ak¦ÿgZv †ei K‡iv| [RUET 19-20] mgvavb: 30 5 m h = 5 sin30 = 2.5 m  †jvKwUi Ak¦ÿgZv, P = mgh t = (60 + 25)  9.8  2.5 10  746 = 0.279 HP 2. GK wgUvi •`‡N© ̈i GKwU mij †`vj‡Ki e‡ei fi 200 g| GUv‡K 60 †Kv‡Y †U‡b †Q‡o w`‡q gy3fv‡e `yj‡Z †`Iqv n‡jv| e‡ei MwZkw3 †ei hLb (i) GUv mvg ̈ve ̄’v w`‡q AwZμg K‡i (ii) myZv j‡¤^i mv‡_ 30 †KvY Drcbœ K‡i (g = 10 ms–2 )| [RUET 18-19] mgvavb: Lcos L–Lcos m = 0.2 kg 60 L (i) mvg ̈ve ̄’vq, Ek = 1 2 mv2 = 1 2 m  2gh = mg(L – Lcos) = 0.2  10(1 – cos60) J = 1 J (Ans.) (ii) E k = ETotal – Ep = 1 – [0.2  10  (1 – cos30)] = 0.732 J (Ans.) 3. m‡e©v”P 1800 kg fi en‡b mÿg GKwU wjdU 2 ms–1 mg‡e‡M Dc‡ii w`‡K DV‡Q| MwZi weiæ‡× Nl©Y e‡ji gvb 4000 N, wjdU Gi Rb ̈ me©wb¤œ KZ HP wewkó gU‡ii cÖ‡qvRb n‡e? [RUET 18-19] mgvavb: T = F + mg = 4000 + (1800  9.8) = 21640 N P = Tv = 21640  2 W = 43280 W = 58.02 HP (Ans.) 4. GKwU Rjwe`y ̈r cvIqvi †÷kb †j‡Ki cvwb e ̈envi K‡i| Uve©vBb †_‡K cvwbi D”PZv 50 m| `ÿZv 50% a‡i 1 MW ÿgZv cvIqv Rb ̈ cÖwZ †m‡K‡Û Uve©vBb w`‡q cÖevwnZ cvwbi fi wbY©q Ki| [RUET 18-19] mgvavb: 50%  mgh = Pt  1 2  m  9.8  50 = 106  1  m = 4081.633 kg (Ans.) 5. 1 m Kvh©Kix •`N© ̈ wewkó GKwU mij †`vj‡Ki e‡ei fi 300 g, †`vjKwU‡K mvg ̈ve ̄’v †_‡K 60 †Kv‡Y wb‡q wM‡q †Qu‡o †`Iqv n‡jv| eewUi MwZkw3 †ei Ki hLb GwU mvg ̈ve ̄’v w`‡q AwZμg K‡i Ges hLb myZv mvg ̈ve ̄’vi mv‡_ 30 †KvY Drcbœ K‡i| [g = 10 ms–2 ] [RUET 18-19, 15-16] mgvavb: mvg ̈ve ̄’vq, Ek = 1 2 mv2 = 1 2 m  2gh = mg(L – L cos) = 0.3  10  (1 – cos60) = 1.5 J (Ans.) E k = ETotal – Ep = 1.5 – [0.3  10  (1 – cos30)] = 1.098 J (Ans.) 6. GKwU cvwb we`y ̈r †K‡›`ai euv‡ai D”PZv 15 m| 5 MW we`y ̈r Drcv`‡bi Rb ̈ cÖwZ †m‡K‡Û UvievB‡bi †eøW ̧‡jvi Dci KZ †KwR cvwb co‡Z n‡e? [RUET 17-18; BUET 10-11] mgvavb: Pt = mgh  m = Pt gh = 5  106  1 9.8  15 = 34.013  103 kg (Ans.) 7. 100 m D”PZv †_‡K 5 kg fi gy3fv‡e AwfK‡l©i Uv‡b co‡Z _vK‡e, 4 sec c‡i fiwUi MwZkw3 I w ̄’wZkw3 KZ n‡e? [RUET 10-11] mgvavb: e ̄‘i †gvU kw3 ETotal = mgH 4 sec ci f‚wg n‡Z e ̄‘i D”PZv, h = H – 1 2 gt2 = 100 – 1 2  9.8  4 2 m = 21.6 m  4 x ci e ̄‘i w ̄’wZkw3 = mgh = 5  9.8  21.6 J = 1058.4 J (Ans.)  4 x e ̄‘i MwZkw3 = mg(H – h) = 5  9.8  78.4 J = 3841.6 J (Ans.) H * h
4 .........................................................................................................................................  Physics 1st Paper Chapter-5 8. 2 kg f‡ii GKwU e ̄‘‡K f‚wg †_‡K Lvov E‡aŸ© wb‡ÿc Kiv n‡jv Ges e ̄‘wU 8 sec ci cybivq wd‡j Gj| wb‡ÿ‡ci gyû‡Z© Ges wb‡ÿ‡ci 2 sec c‡i e ̄‘wUi wefe kw3 Ges MwZ kw3 KZ? (g = 9.8 ms–2 ) [RUET 09-10] mgvavb: T = 2u g  u = 8  9.8  0.5  u = 39.2 ms–1 wb‡ÿ‡ci gyn~‡Z© wefekw3 = 0 J (Ans.) MwZkw3 = 1 2 mu 2 = 1536.64 J (Ans.) 2 s ci, v = u – 2g = 19.6 ms–1  2 s ci MwZkw3 = 1 2  2  19.62 = 384.16 J (Ans.) wefekw3 = (0 + 1536.64) – 384.16 J = 1152.48 J (Ans.) 9. †Kv‡bv Kzqv †_‡K 20 m Dc‡i cvwb †Zvjvi Rb ̈ 6 kW ÿgZvi GKwU cv¤ú e ̈envi Kiv n‡”Q| cv‡¤úi `ÿZv 82.2% n‡j cÖwZ wgwb‡U KZ wjUvi cvwb †Zvjv hv‡e? [RUET 08-09] mgvavb: P = mgh t  6  103  0.822 = 103  V  9.8  20 60  V = 1.5 m3 (Ans.) 10. 25 gm f‡ii GKwU ey‡jU 100 cms–1 †e‡M 15 cm cyiæ GKwU Kv‡Vi †`qv‡j cÖ‡ek K‡i I †`Iqvj †f` K‡i 75 cms–1 †e‡M †ewi‡q hvq| ey‡j‡Ui Mo ej KZ? [RUET 08-09, 07-08] mgvavb: F = ma = m v 2 – u 2 2s = 25  10–3 0.752 – 1 2 2  0.15 N = – 0.036 N (Ans.) 11. 60 kg fiwewkó GKwU e ̄‘ w ̄’i Ae ̄’vq wQj| 30 N ej cÖ‡qvM Kivq e ̄‘wU MwZcÖvß n‡jv| 10s c‡i e ̄‘wUi MwZkw3 wbY©q K‡iv| [RUET 07-08] mgvavb: 10 s ci e ̄‘wUi †eM, v = u + at = u + F m t = 0 + 30 60  10 = 0.5  10 = 5 ms–1  10 s ci e ̄‘wUi MwZkw3, Ek = 1 2 mv2 = 1 2  60  5 2 = 30  25 = 750 J (Ans.) 12. GKwU ̧wj cÖwZ †m‡K‡Û 200 wgUvi mij MwZ‡Z P‡j 50 cm cyiæ GKwU Kv‡Vi ̧uwo‡K †Kv‡bv iK‡g †Q` K‡i| H GKB ai‡bi ̧wj GKB Kv‡Vi 40 cm cyiæ ̧uwo n‡Z KZ †eM †ei n‡e? [RUET 06-07] mgvavb: a = u 2 – v 2 2s = 2002 2  0.5 ms–2 = 40000 ms–2 v = u 2 – 2as = (200) 2 – (2  4000  0.4) = 89.44 ms–1 (Ans.) 13. GKwU cv¤ú wgwb‡U 1200 gallon cwigvY cvwb 6 ft DuPz‡Z 32 ft–1 (9.8 ms–1 ) MwZ‡e‡M wb‡ÿc Ki‡Z cv‡i| 1 gallon cvwbi fi 10 lb n‡j Bwćbi Ak¦ÿgZv wbY©q K‡iv| [RUET 06-07, 05-06] mgvavb: 1 lb = 0.4536 kg 1 ft = 0.3048 m  m = 1200  10  0.4536 kg = 5443.2 kg h = 6  0.3048 m = 1.83 m  W = mgh + 1 2 mv2 = (5443.2  9.8  1.83) + {}      1  2  5443.2  9.82 = 3.59  105 J  P = W t = 3.59  105 60  746 HP = 8.02 HP (Ans.) 14. 70 kg f‡ii GK e ̈w3 20 kg f‡ii GK †evSv wb‡q 6 m `xN© GKwU wmuwo †e‡q Dc‡i DV‡jv| wmuwowU Avbyf‚wgK Z‡ji mv‡_ 30 †KvY K‡i _vK‡j H e ̈w3 KZ KvR Kij wbY©q K‡iv| [RUET 04-05] mgvavb: B A C AB = 6 m 30 W = mgh = mgs sin = (70 + 20)  9.8  6  sin30 = 2646 N (Ans.)