Q#2 (a) You are driving across a gravel bed, when you see apatch of ice ahead, y
ID: 1674814 • Letter: Q
Question
Q#2 (a) You are driving across a gravel bed, when you see apatch of ice ahead, you salm on the brakes attempting to stopbefore you hit the ice. The kinetic energy dissipated in thebraking maneuver was 3.0 * 10 to powe 5 J . The maneuverbrings your vehicle (assume the vehicle mass is 1.0 *10 to powe 3kg) to rest at the edge of the ice patch. what was your initialvelocity? (b) In attempting to turn your car arround, you inadvertently end upstuck on the patch of ice ( assume it is africtionless surface) and have be pulled off the surfaceby a tow truck with a cable and winch. The tow truck cable is at anangle of 40.0 degrere with respet to the horizontal surface. Thetention in the cable during the pulling orocess is 3.50 *10to power3 N. The car has a mass of 1.0 *10 to power 3 kg. How much work isdone by the cable in moving the car a distance of 15 m?. (c) Thetow truck driver pulls your vehicle onto the flat- bed tow truck bypulling it up onto the bed of the truck. what is the change in thegravitational potential energy of the car after it is loaded ontothe truck and at a height of 1.8 m above the ground?. Q#2 (a) You are driving across a gravel bed, when you see apatch of ice ahead, you salm on the brakes attempting to stopbefore you hit the ice. The kinetic energy dissipated in thebraking maneuver was 3.0 * 10 to powe 5 J . The maneuverbrings your vehicle (assume the vehicle mass is 1.0 *10 to powe 3kg) to rest at the edge of the ice patch. what was your initialvelocity? (b) In attempting to turn your car arround, you inadvertently end upstuck on the patch of ice ( assume it is africtionless surface) and have be pulled off the surfaceby a tow truck with a cable and winch. The tow truck cable is at anangle of 40.0 degrere with respet to the horizontal surface. Thetention in the cable during the pulling orocess is 3.50 *10to power3 N. The car has a mass of 1.0 *10 to power 3 kg. How much work isdone by the cable in moving the car a distance of 15 m?. (c) Thetow truck driver pulls your vehicle onto the flat- bed tow truck bypulling it up onto the bed of the truck. what is the change in thegravitational potential energy of the car after it is loaded ontothe truck and at a height of 1.8 m above the ground?.Explanation / Answer
a) m=1.0x103 kg W=3.0x105 J Since I hope th eenergy was conceved Ke= W Ke=0.5mV2 then V= (2Ke /m)=(2W /m)= V = (2 x3.0x105 /1.0x103 )= V= 24.5 m/s b) d=15.0 m F= 3.50 x10 3 N angle 40 deg with respect tohorizontal Work W is by definition W= F d cos () W= 3.50 x10 3 x 15 x cos(40) W= 40,200J c) h=1.8 m=1.0x103 kg Potential energy Pe is Pe= mgh Pe= 1.0x103 x 9.8 x 1.8 = Pe= 17,640 J Pe= 17,600 JRelated Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.