Question

The Scalar Product The vectors A and B are given by A = 1 + 3j and B = -1i + 13 (a) Determine the scalar product AB. SOLUTION Conceptualize There is no physical system to imagine here. Rather, it is purely a mathematical exercise involving two-Select- Categorize Because we have a definition for the scalar product, we categorize this example as --Select problem. Substitute the specific vector expressions for A and B A B - (11 + 3) - (-17 + 11) + + + The same result is obtained when we use the equation A-8-4,8, + A,B,+A_, directly, where 4 - 1,4, -3,8, --1 and 8, - 1. (b) Find the angle between A and B. SOLUTION Evaluate the magnitudes of A and using the Pythagorean theorem: A. VAA, B+ Use the equation AB - AB cos(6) and the result from part (a) to find the angle (in degrees). cos(0) - AB 8. Cos1 AB EXERCISE As a particle moves from the origin to (98 + 9), it is acted upon by a force given by (94 + 7) N. Calculate the work done by this force on the particle as it moves through the given displacement. (Give your answer in J.) Hint W-

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Answer 1

6-35), (-11-11) [itEO2) + [it anjJ+ [30--02]-(33-15) 33- - - +0-300+3 AB - २

W 75N J210 fix <ingh displacement d=5-7m Mk o.150 workdone on block by Exercise mass of block - 15 kg. 75cosai wang F = 755 2 1 - fx = 75 casai - MKN = 75 cosajo – 0.150x1589.8 F = 47.96 normal force N. work done by the 75 force => (47.96) (5.7) 273.37 Joule 2 w = Fdcosao the gravitational work done on block by w=fdwas ao force .

너 Increase in internal energy of the block - surface system due to friction 75 85 # 47.96 457 ou 154.28 Joule e Change in kinetic Energy Ks-K, = w 273.37 Joule OK