Asked by jim
A 5.56 kg block located on a horizontal floor is pulled by a cord that exerts a force F = 10.7 N at an angle theta = 17.0° above the horizontal, as shown in the Figure. The coefficient of kinetic friction between the block and the floor is 0.06. What is the speed of the block 6.1 s after it starts moving?
Answers
Answered by
Henry
Wb = m*g = 5.56kg * 9.8N/kg = 54.5 N. =
Wt. of block
Fb = 54.5N @ 0o. = Force of the block.
Fp = 54.5*sin(0) = 0. Force parallel to
floor.
Fv = 54.5*cos(0) - 10.7*sin17 =51.4 N. =
Force perpendicular to floor.
Fk = u*Fv = 0.06 * 51.4 = 3.08 N.=Force
of kinetic friction.
Fap-Fp-Fk = ma.
10.7*cos17-0-3.08 = 5.56*a
10.5-3.08 = 5.56a
7.43 = 5.56a
a = 1.34 m/s^2.
V = Vo + at = 0 + 1.34*6.1 = 8.16 m/s.
Wt. of block
Fb = 54.5N @ 0o. = Force of the block.
Fp = 54.5*sin(0) = 0. Force parallel to
floor.
Fv = 54.5*cos(0) - 10.7*sin17 =51.4 N. =
Force perpendicular to floor.
Fk = u*Fv = 0.06 * 51.4 = 3.08 N.=Force
of kinetic friction.
Fap-Fp-Fk = ma.
10.7*cos17-0-3.08 = 5.56*a
10.5-3.08 = 5.56a
7.43 = 5.56a
a = 1.34 m/s^2.
V = Vo + at = 0 + 1.34*6.1 = 8.16 m/s.
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