Question

I need help with question 6 and 8, please. Thanks so much! :)

wa 3 the 5. The combined weight of a motorboat and its load is 480 egu pounds. After starting from rest at t= 0, the motor exerts a dow constant force of 30 pounds on the boat in its linear direction of of s motion. If the total resistance (in pounds) that the boat encounters (bu as it moves along its path is numerically equal to 3v12, where vis the the velocity (in feet/sec) of the boat, what is the velocity of the boat at the end of 20 seconds? 6. A loaded sled is pulled across a level field of ice with a An constant speed of 15 feet/sec. When the sled is halfway across 1. the ice field (i.e., at t = 0) the sled is released and it continues on its own. The air resistance in pounds) is numerically equal to 2. the sled's velocity (in feet/sec). If the weight of the loaded sled is 80 pounds, and if the coefficient of sliding friction of the sled's runners on the ice is 1/20, how far will the sled continue to move on its own after it is released? 7. Starting from rest at t = 0, an 8 pound object is pulled along a horizontal surface with a force (in pounds) that is equal to twice the distance in feet) of the object from its starting point x = 0. If the air resistance (in pounds) is numerically equal to 1218, where v is the velocity (in feet/sec) of the object, and if the coefficient of sliding friction is 1/4, (a) what is the velocity of the object in terms of the distance x (in feet) that the object has traveled, and (b) how far has the object gone when its velocity is 4 feet/sec? Hint: after setting up the DE that solves this problem, you will need to make the substitution y=v2 in your DE in order to convert it to a DE that you will know how to solve. 8. The combined weight of a boy and his sled is 100 pounds. Starting from rest at t= 0, he sleds down a hill whose slope is 74. The air resistance (in pounds) is numerically equal to twice the velocity (in feet/sec) of the boy and his sled. If the coefficient of sliding friction between the sled's runners and the sloping hill is 1/10, (a) what is the velocity of the sled at the end oft seconds, and (b) how long (correct to the nearest foot) is the hill if it take 5 seconds to reach the bottom of the hill?

Answer 1

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For the detailed solution please find the attached image below

u = 15ft/s -Fa J = 32 ft/s2 2 ܘܢ fie NET FORCE OF RESISTANCE ON BLOCK RESISTANCE f FRICTION = AIR = Thet V + Mk mg Ind LAW, BY NEWTON'S I vt free 20 ) 4) (8 ma TT 80 be 32 32 32 ft/s2 +1 20 a = 80

- dv dx = ( (% +) 32 .Olv ده 31 - هری v tl o 0 ) 20 کا بنا ه-د32 - 3 2 [ + ] ما ما 3 - 3ov] لا )ک - 3 2 1 -- 707.39 2 - 70 1039 م 2/09 - 3 )

VEUX ostante 8. N v Fa fie = 37° masino mg coso SO G 5 = Efyo myco coso g=32 ft/s2 N=ng cos = (100)@) NET FORCE ALONG THE SLED 80 lb ft/s IFNET FNET masino - Fa = fx ng sino 2V - HK N (100) 30 - 2V -H. ( (80) = 5 2 - 2V . 99 SADES El

= a= FNET (0:52-)32 te E 52 -2V 100 32 m du = If du - .32 > (0.52 - **)32 (0:52-5) that -50 en ( 26-~) S E 32 t ņ ņ 뼱 9) 26-v -16 t 26 -16t v= 26 - 26 e 25 dx at -16% 5- 26 e - 167 26 e 25 Now V= 26 C -).olt 25 38(260 - J 16t 25 s dx - (0-0) 22 -> - [267 – 201-25). e p = 91 ft b) LENGTH OF 8 HILL Six=