Posted by B on .
At the beginning of a new school term, a student moves a box of books by attaching a rope to the box and pulling with a force of 92 N at an angle of 30 as shown in the figure below. The box of books has a mass of 22.0 kg, and the coefficient of kinetic friction between the bottom of the box and floor is 0.5.
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..
Wb = 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
b. V ^2 = Vo^2 + 2a*d