Three forces acting on an object are given by vector F1 = ( - 2.20 i hat + 2.10

Question

Three forces acting on an object are given by vector F1 = ( - 2.20 i hat + 2.10 j ) N, vector F2 = ( 4.80 i hat + 3.00 j) N, and vector F3 = ( 45.5 i hat + 50.0 j) N. The object experiences an acceleration of magnitude 3.55 m/s2.

(a) What is the direction of the acceleration?

_____° (from the positive x axis)

(b) What is the mass of the object?_________


(c) If the object is initially at rest, what is its speed after 19.0 s?________

(d) What are the velocity components of the object after 19.0 s?
(__________ i +___________ j )

Please state the answer, units, and how you got it. Double check as well to make sure its correct. Thanks

Explanation / Answer

let,

F1=(-2.2)i+(2.1)j N

F2=(4.8)i+(3)j N

F3=(45.5)i+(50)j N

acceleration, a=3.55 m/sec^2

a)

along x-axis, Fx=(-2.2+4.8+45.5) = 48.1 N

along y-axis, Fy=(2.1+3+50) = 55.1 N

use,

tan(theta)=Fy/Fx

tan(theta)=55.1/48.1

====> theta=48.9 degrees or 49 degrees

b)

Fnet=m*a

m=Fnet/a

m=sqrt(48.1^2+55.1^2)/3.55

m=20.6 kg

mass m=20.6 kg

c)

at t=19 sec

use,

V=u+a*t

V=0+3.55*19

V=67.45 m/sec

d)

vx=v*cos(theta) =67.45*cos(48.9)=44.34 m/sec

vy=v*sin(theta)=67.45*sin(48.9)=50.83 m/sec

and

v=(44.34)i+(50.83)j

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