CALC? Starting at t = 0, a horizontal net force is applied to a box that has an initial momentum What is the momentum of the box at t = 2.00 s?
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Solution 16E Momentum ( p ) and variable force ( F) are related by the following equation, t2 p = Fdt t1 Given that, net force F = (0.280 N/s) t i + ( 0.450 N/s )t j Initial momentum p = ( 3.00 kg. m/s) i + (4.00 kg.m/s) j Let us now calculate the x-component first. 2 p x (0280 N/s)t i dt 0 t2 2 2 p x 0.280 N/s i ( 2 s0 p = 0.560 N.s i x Let us now calculate the y-component. 2 p y (0.450 N/s )t jdt 0 2 t3 2 p y 0.450 N/s j ( ) 3 0 p = 0.150 N/s j × 8 s 3 y p y 1.2 N.s j Therefore, the net x-component of momentum = 3.00 kg.m/s i + 0.560 kg.m/s i = 2.44 kg.m/s i The net y-component of momentum = 4.00 kg.m/s j 1.2 kg.m/s j = 2.8 kg.m/s j Therefore, omentum at t = 2.00 s ( 2.44 kg.m/s i + 2.8 kg.m/s j)
Textbook: University Physics
Author: Hugh D. Young, Roger A. Freedman
University Physics was written by and is associated to the ISBN: 9780321675460. The answer to “CALC? Starting at t = 0, a horizontal net force is applied to a box that has an initial momentum What is the momentum of the box at t = 2.00 s?” is broken down into a number of easy to follow steps, and 32 words. This textbook survival guide was created for the textbook: University Physics, edition: 13. Since the solution to 16E from 8 chapter was answered, more than 925 students have viewed the full step-by-step answer. The full step-by-step solution to problem: 16E from chapter: 8 was answered by , our top Physics solution expert on 05/06/17, 06:07PM. This full solution covers the following key subjects: box, momentum, horizontal, Force, Calc. This expansive textbook survival guide covers 26 chapters, and 2929 solutions.