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September 21, 2014

September 21, 2014

Posted by **B** on Monday, November 5, 2012 at 2:48pm.

a) Find the acceleration of the box.

b) If the box is at rest to start with, what is its speed after it has traveled 2.00 m?

c) How much time does it take to pull the box this distance?

Please show how you got there! This problem has stumped me..

- Physics -
**Henry**, Wednesday, November 7, 2012 at 5:23pmWb = m*g = 22kg * 9.8N/kg = 215.6 N. =

Wt. of box.

Fb = 215.6N @ 0o. = = Force of box.

Fp = 215.6*sin(0) = 0 = Force parallel to floor.

Fv = 215.6*cos(0) - 92*sin30 = 169.6 =

Force perpendicular to floor=The Normal.

Fk = u*Fv = 0.5 * 169.6=84.8 N. = Force

of kinetic friction.

a. Fap*cos30-Fp-Fk = m*a.

92*cos30-0-84.8 = 22a

22a = -5.12

a = -0.23 m/s^2. This should not be

negative. Make sure all given INFO is

correct.

b. V ^2 = Vo^2 + 2a*d

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