Question

2 55 pes) Use work energy theorem to solve 1650 kg) is at the soap of a menge there to solve this problem. A rocket powered sled ( pot a ll that is angled at 28. that is angleda alsave the horizontal. The coefficient concc friction between the constant chest of 175 and the sledis 0.136. The rockets on the sled provide a and the stad is 0.136. The rockets on SR The sled travels a da The sled travels a distance of 21.5 m to reach the bottom of the ... A) Find the work is done by each force acting on the sle ) moving at the bottom of the How would speed of the sted be diere 9 of the hill was smaller than 28.2° if the sted meves 21.5 m down the hills each torce acting on the sled. B) How fast is the sled

Answer 1

Son: m-1650 KS 28.2 ,HeO-136, Fe 15 N,d- 21.5m JE y Sin ce mg coso don byoucet, coFd = 195*21.5 3162.5J CooTk done by Norsmal to, CUN = 0 CooBk done by avity isCoj = mg Sino d - 16 50% q8 8m(29-2 ") x21-5 Cu164284. 64J CooK done bychal or is =-4mg Cosa dl =-0-18671650xq.21. S Cos( Foora FBD Ftmg Snu-d Ft mg Sino hma Mengcore Cosa= hna 15+ 165ox9.8x Sin (282- 0136 1650xq.8x Cos(28-2) = 1650 a 1650a- 7816.(46-1938 092 a 3.562 hmls v2u2-2ad (u-0) V2ad 213-56221-S V 12.311 mg4 Speed at the bollom hill.

Agai m FBD Ft ma Sma-ly mg coso ma 16170 Sine-2199-12 Coso= 1650 a a 0.106+ 9.8 Sno-1.333 Coso 28.2 then ac8.562 m2 So, V< 12.317ml ge, Speed sled totll be dferent angle oas Smaller than 28.2