This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: patel (kp9583) HW4 Mackie (20211) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points The position of a toy locomotive moving on a straight track along the x-axis is given by the equation x = t 5- 6 t 2 + 9 t , where x is in meters and t is in seconds. The net force on the locomotive is equal to zero when t is equal to Correct answer: 0 . 843433 s. Explanation: From Newtons second law of motion, the force is directly proportional to the accelera- tion, which means zero acceleration will give a zero force. The velocity is v = d x dt = 5 t 4- 12 t and the acceleration is a = d v dt = 20 t 3- 12 , so a = 0 will give t = 0 . 843433 s s. 002 10.0 points A loaded truck can accelerate at 4 . 7 m / s 2 . It loses its load so that it is only 0 . 6 as massive. By what factor must the acceleration change for the same driving force? Correct answer: 1 . 66667. Explanation: F net = ma Since the mass changes by a factor of n , and the force does not change, multiply the right side by a clever form of 1 = n 1 n : F net = ma = ( nm ) parenleftBig 1 n a parenrightBig and the acceleration must change by a factor of 1 n . 003 (part 1 of 2) 10.0 points A child holds a sled on a frictionless, snow- covered hill, inclined at an angle of 27 . 8 7 N F 27 If the sled weighs 87 N, find the force ex- erted on the rope by the child. Correct answer: 39 . 4972 N. Explanation: Given : W = 87 N and = 27 . Consider the free body diagram for the block m g s i n N = m g c o s F W Basic Concepts: If we tilt our world, and consider the forces parallel to the hill, F net = summationdisplay F up- summationdisplay F down = 0 then the forces perpendicular to the hill, F net = summationdisplay F out- summationdisplay F in = 0 Solution: Consider the free body diagram for the sled: The weight of the sled has compo- nents W sin acting down the hill and W cos acting straight into the hill. The system is in equilibrium, so for forces parallel to the hill, F net = T-W sin = 0 patel (kp9583) HW4 Mackie (20211) 2 = T = W sin = (87 N) sin27 = 39 . 4972 N 004 (part 2 of 2) 10.0 points What force is exerted on the sled by the hill? Correct answer: 77 . 5176 N. Explanation: For forces perpendicular to the hill, F net = N -W cos = 0 = N = W cos = (87 N) cos27 = 77 . 5176 N 005 (part 1 of 2) 10.0 points A block of mass 6 . 71 kg lies on a frictionless horizontal surface. The block is connected by a cord passing over a pulley to another block of mass 5 . 38 kg which hangs in the air, as shown on the following picture. Assume the cord to be light (massless and weightless) and unstretchable and the pulley to have no friction and no rotational inertia....
View Full Document
This note was uploaded on 12/10/2009 for the course PHYSICS 2021 taught by Professor Mihalisin during the Fall '09 term at Temple.
- Fall '09