Wednesday

April 16, 2014

April 16, 2014

Posted by **anonymous** on Wednesday, March 3, 2010 at 5:51am.

F3 = ( - 45.0i + 48.0j ) N. The object experiences an acceleration of magnitude 3.75 m/s2.

(a) What is the direction of the acceleration?

_________° (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 after 17.0 s?

_________ m/s

(d) What are the velocity components of the object after 17.0 s?

( ________ i + _________ j ) m/s

- physics -
**drwls**, Wednesday, March 3, 2010 at 6:45amThis is an exercise in adding vectors. The ratio of components in vector sum will provide the fdirection information you need. The mass is the ratio of the force magnitude to the acceleration.

For (c), assume the initial speed is zero and multiply the acceleration and time interval. For (d), multiply the vector acceleration by the time.

Show your work if further assistance is needed.

- physics -
**Anonymous**, Tuesday, October 12, 2010 at 8:38pma) direction of acceleration in the same way as resultant force, hence

Fr=F1+F2+F3=(-1.85-5.10-45.0)i+(2.50+3.25+48.0)j= -51.95i+53.75j [N]

θ=tanˉ¹(53.75/-51.95)=134°.

b) F=ma --- m=F/a --- (√((-51.95)^2+53.75^2))/3.75=19.9kg

c)Vf=Vi+at=0+3.75*17=63.75m/s

d)F=ma --- a=F/m ---āx=F̅x/m --āy=F̅y/m

ā=āx + āy

V̅f=V̅i+āt=0+((-51.95/19.9)i+(53.75/19.5)j)*10

V̅f=((-51.95/19.9)*10)i+(53.7/19.5)*10)j

V̅f= -26.1i+27.5j

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