Three forces acting on an object are given by Fi magnitude 3.55 m/s2. (-1.95î +7

Question

Three forces acting on an object are given by Fi magnitude 3.55 m/s2. (-1.95î +7.40]) N, F2 - (5.15î - 1.3j) N, and F3 -(-47î) N. The object experiences an acceleration of 1.3) N, and F -473) N. The object (a) What is the direction of the acceleration? 7.8 Note that the direction of the acceleration is the same as the direction of the net force. (counterclockwise from the +x-axis) (b) What is the mass of the object? kg (c) If the object is initially at rest, what is its speed after 15.0 s? 53.25 (d) What are the velocity components of the object after 15.0 s? (Let the velocity be denoted by V.) v= j m/s

Explanation / Answer

a) Fnet = F1 + F2 + F3

        = (-1.95 + 5.15 -47)i + (7.40 -1.3)j

        = -43.8i + 6.1j

Fnet = ma

so direction of Fnet and a is same .

hence

@ = 180 - tan^-1(6.1/43.8) = 172.07 deg

b) |F| = m |a|

|F| = sqrt(43.8^2 + 6.1^2) = 44.22 N

|a| = 3.55 m/s^2

m = |F| / |a| = 12.46 kg

c) using , v = u + at

v = 0 + (3.55 x 15) = 53.25 m/s

d) vector a = F / m = -43.8i + 6.1j / 12.46

a = - 3.52i + 0.49j   m/s^2

v = u + at

v = 0 + (- 3.52i + 0.49j )15

v =-52.8i + 7.34j

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