Suppose we pull a 3.0-kg block of material along a level surface, using a string
ID: 2108972 • Letter: S
Question
Suppose we pull a 3.0-kg block of material along a level surface, using a string inclined at an angle θ = 25 Degrees above the horizontal. The tension in the string is 18.0 N. The coefficient of kinetic friction between block and surface is 0.25.
(A) Draw a free-body diagram of the block. There should be four arrows representing force vectors.
(B) Determine the normal force on the block.
(C) Determine the force of friction (5.45 N opposite the velocity)
(D) Find the acceleration of the block (3.6 m/s2 .)
(E) Suppose we keep the tension constant, but vary the angle from zero through 85 Degrees. At what angle will the acceleration be the greatest? If you decide to use a "brute force: method, a spreadsheet will make things much easier.
Explanation / Answer
Suppose we pull a 3.0-kg block of material along a level surface, using a string inclined at an angle Î ¸Ã‚ = 25 Degrees above the horizontal. The tension in the string is 18.0 N. The coefficient of kinetic friction between block and surface is 0.25.
(A) Draw a free-body diagram of the block. There should be four arrows representing force vectors.
(B) Determine the normal force on the block.
N = m g - T sin(Theta) = 3*9.8 - 18 * 25s = 21.793 = 21.8 N
(C) Determine the force of friction  (5.45 N opposite the velocity)
fk = u N = 0.25 * 21.793 = 5.4482 = 5.45 N
(D) Find the acceleration of the block (3.6 m/s2 .)
a = (T cos25 - fk)/m = (18*25o - 5.4482)/3 = 3.6218 = 3.62 m/s2
(E)Suppose we keep the tension constant, but vary the angle from zero through 85 Degrees. At what angle will the acceleration be the greatest? If you decide to use a "brute force: method, a spreadsheet will make things much easier.
a = (T cos(t) - fk)/m = (T cos(t) - u N)/m = (T cos(t) - u (m g - T sin(t)))/m
==> da/dt = (-T sin(t) - u m g + u T cos(t))/m
==> da/dt = 0
==> -T sin(t) - u m g + u T cos(t) = 0
==> -18*sin(t) - 0.25*3*9.8 + 0.25*18*cos(t) = 0
==> t = 14.3 degrees