Question

Require three significant digits

5.2kg object, initially at rest, is acted on by a net external force in a straight line described by F = 14 (1 - exp(-0.21))N, where tis in seconds. (a) What is the magnitude of the momentum of the object after 10s have elapsed? Number Units (b) What is the work of the net external force after 10s have elapsed? Number Units

Answer 1

Finding momentum and work done from an expression of net force

m z 5.2 kg F 214(1-02+) N (a) If þ be the momentum dp, FF or dp z 14(1- ě o2t) p z linci-e 0.2 t ) dt NNN - 0 2+ ) 4t -0.2t dt -0.2t7 to + z 14XL -02 to 2 14 lat - 14 e 2.14x [+7.-19* [6021 2.14x10 + 12x (é 2-1) 2.140 + 70 (-0.864664) 2 79.5 kgm/s (rounded to 3 significant digits) - 6.2 o 2% t . e or con on dt 5.2 d x = 14 (1- ) du 2.21692307 (1-e v 2. 2:692307 16-2-02t) dt 22.69 2307(t - CO2 ) + - where c is an -oztarbitrary constant. -0.2t tc + c 22692307 t + 13'4615 35e

As the p object starts from rest 0 = 13.461535 + C on C 2-13:461535 - 0 2t -13'4615 35 - V2 2.692307 t + 13.461 535 ... kower ,P z Fun 214(1 - 8 024) Ⓡ (2.692307 + + 13.46 1535 en 13.46 1535 z 37.692298 +- 37 69 2298 te 2+168461491 -0.4t - 188.46199 e 18846149 + 188'461.49 en - 0 2t ? 37.692298 t - 37.69 22:98 te +376.92298e -0.4t 18646149 e 18.8.46149 ... work done a Pdt.. -02t te at 2 37 692 2.98 t at 37.69229 10 ,10 -0.4t dt + 376.92298 6-0 2t dt - 188 46149 e -188.46149 dt .... ) stat a ↑ [+ 2] 10 2 50 10 te025 -025 710 te dt z It e -0. 2t L - 0:2 do J-02 -0-24710 2-6.2 x 11355352 + o'z x 02 -0.2 Jo 2 14.84985 2 - 6.76676 -

- 0.2 + idt z -0.2t 710 2 L - 0.2 2 4323323 10 -0.9t dt. z. - 0.4t yo l L-04 dt. z. I en 2- oa (€ 9-21), 2:45 42109 s at a [+] !" 2 10. From (1) work done z 37 692298 50- 37.69 2298 X 14.84985 + 376 92298 x 4.323323-18 7:46149 X 2'454 2109 168.461 4.9 X 10 2 608 J . (rounded to 3 significant A shorter method for solution of (b) is to use uz p in work energy theorem: W z I mur - 0 2 1 / 2 mg .. which I figured out later , after I a zoved the problem in the longer method. shown above.