The same result is obtained when we use the equation A middot B = A, B + AQ, B +

Question

The same result is obtained when we use the equation A middot B = A, B + AQ, B + A, B directly, where A_x = 5, A_y = 5, B_x = -1, and B_y = 7. (B) Find the angle theta between A and B. Evaluate the magnitudes of A and B using the pythagorean theorem: A = Squareroot A^2_x + A^2_y = Squareroot (5)^2 + (5)^2 = Squareroot 50 B = Squareroot B^2_x + B_y^2 = Squareroot (-1)^2 + (7)^2 = Squareroot 50 Use the equation A middot AB cos (theta) and the result from part (A) to find the angle: cos (theta) = A middot B/AB = A middot B/Squareroot 50 Squareroot 50 = A middot B/Squareroot 2500 Theta = cos^-1 A middot B/Squareroot 2500 theta = As a particle moves from the origin to (3i + 8j) m, it is acted upon by a force given by (Si + 1j) N. Calculate the work done by this force on the particle as it moves through the given displacement.

Explanation / Answer

Part 1)

vector A=sqrt(50)

vector B= sqrt(50)

Theta= inv(cos) A.B/(sqrt(2500))

=inv(cos)50/50

=inv(cos)1

=0 degrees

Ans- Theta= 0 degrees

Part 2)

d=(3i+8j) m

F= (8i+1j) N

W=f.d

=24+8

=32 J

Ans- Work done by the force is 32 J

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