answers please Three forces acting on an object are given by F 1 = (1.75 i - 2.5

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answers please
Three forces acting on an object are given by F1 = (1.75 i - 2.50 j ) N, F2 = ( 5.00 i + 3.35 j) N, andF3 = ( - 41.0 i ) N. Theobject experiences an acceleration of magnitude 3.75 m/s2. (a) What is the direction of theacceleration?
_____________° (from the positive x axis)

(b) What is the mass of the object?
______________ kg

(c) If the object is initially at rest, what is its speed after20.0 s?
_______________ m/s

(d) What are the velocity components of the object after20.0 s?
( 4i + 5j ) m/s answers please
Three forces acting on an object are given by F1 = (1.75 i - 2.50 j ) N, F2 = ( 5.00 i + 3.35 j) N, andF3 = ( - 41.0 i ) N. Theobject experiences an acceleration of magnitude 3.75 m/s2. (a) What is the direction of theacceleration?
_____________° (from the positive x axis)

(b) What is the mass of the object?
______________ kg

(c) If the object is initially at rest, what is its speed after20.0 s?
_______________ m/s

(d) What are the velocity components of the object after20.0 s?
( 4i + 5j ) m/s

Explanation / Answer

Given F1 = ( 1.75 i - 2.50 j )N
F2 = ( 5.00 i +3.35 j) N F3 = ( - 41.0 i ) N Net force F = F1 + F2 + F3                    =(-34.25 i + 0.85 j ) N acceleration a = 3.75m/s2 Let the net force makes an angle with neagtivre x-axis then tan = 0.85/34.25                                                                                                    = 1.421degrees Therefore direction of force with positive x- axis = 180 - = 178.57 degrees Let the net force makes an angle with neagtivre x-axis then tan = 0.85/34.25                                                                                                    = 1.421degrees Therefore direction of force with positive x- axis = 180 - = 178.57 degrees (a) the direction of the acceleration =dirction of net force                                                     = 178.57 degrees with positive x-axis
(b) the mass of the object m = magnitude of net force /magnitude of accleration
                                            = { [ [ -34.25^ 2 + 0.85^ 2] ] / 3.75 }                                             = 34.26/3.75                                             = 9.136 kg

(c) initial velocity u = 0 time t = 20.0s velcotiy after time t is v = u + at                                     = 0 + [(F / m ) * 20 ]                                     = [(-34.75i +0.85 j ) / 9.36 ] * 20                                     = [-3.803 i + 0.0908 j] * 20                                     = -76.12 i +1.816 j Speed after 20 s is v = [ -76.12^2 + 1.816^ 2]                                 = 76.14 m / s (d). the velocity components of the objectafter 20.0 s  is v = -76.12 i +1.816 j                                 

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