A group of people is pulling on a rope with a force of magnitude 95.0 N. The rop
ID: 1783833 • Letter: A
Question
A group of people is pulling on a rope with a force of magnitude 95.0 N. The rope passes through a frictionless pulley and is attached to a 10.0-kg box that moves up an inclined plane (20 degrees). The box has a speed of 1.50 m/s when the group begins to pull, and the box has pulled a distance of 5.00 m. The coefficient of kinetic friction between the box and the floor is 0.400.
Determine the work done on the box by (a) the force of gravity, (b) the force of friction, (c) the tension in the rope, and (d) the normal force from the inclined plane. (e) Find the change in the kinetic energy of the box. (f) What is the speed of the box after being pulled 5.00 meters? To answer parts (e) and (f), think of the Work-Energy Theorem
Explanation / Answer
a)
gravitational force component acting down the incline Fg = m*g*sintheta
work done by gravity Wg = Fg*s = m*g*sintheta*s*cos180 = -10*9.8*sin20*5 = -167.6 J
(b)
frictional force fk = uk*m*g*costheta
work done by frictioanl force wf = fk*s*cos180 = -0.4*10*9.8*cos20*5 = -184.2 J
(c)
work dons by tension WT = T*s*cos0 = 95*5 = 475 J
(d)
normal force is perpendicular
work done by normal force Wn = n*s*cos90 = 0
(e)
work energy theorem
total work = change in kinetic energy
change in KE = Wg + Wf + WT + Wn
change in KE = -167.6 - 184.2 + 475 + 0 = 123.2 J
================
(f)
dKE = W
(1/2)*m*(vf^2 - vi^2) = W
(1/2)*10*(vf^2-1.5^2) = 123.2
vf = 5.18 m/s <<<-------------ANSWER
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