Answer #2 by showing ALL your work. Make sure to put units in each step. Don’t f
ID: 1792419 • Letter: A
Question
Answer #2 by showing ALL your work. Make sure to put units in each step. Don’t forget to circle the answer. Be sure that I can see al the work and it’s not cut off when the picture is uploaded.
2. (15 points) A mover slides a dresser up a 35° ramp to the loading dock 2 m above the ground. The dresser moves at a constant speed of 0.5 m/sec and the mass of the dresser totals 50 kg. There is a coefficient of friction of 0.20 between the dresser and the ramp. (a) What is the total work done on the dresser? (b) What is the force exerted by the mover? (c) What is the work done by gravity? (d) What is the work done by the normal force? (e) What is the work done by friction?Explanation / Answer
2)
(a)
total work = change in KE
work = (1/2)*m*(vf^2 - vi^2)
given initial speed = final speed = 0.5 m/s
work = 0 <<<<---------------ANSWER
==============================
b)
acceleration a = rate of change in speed = 0
from newtons second law
Fnet = m*a
F - Fg - fk = 0
F = Fg + fk
F = m*g*sintheta + uk*m*g*costheta
F = (50*9.8*sin35) + (0.2*50*9.8*cos35)
F = 361.33 N <<<<<<------ANSWER
====================
(c)
work done by gravity
Wg = -dU = ui - uf
Wg = 0 - m*g*h
Wg = -50*9.8*2
Wg = -980 J <<<<<<------ANSWER
===================
d)
work done by friction
Wf = fk*L*cos180
Wf = -uk*m*g*cos35*L*cos180
L = length of the ramp = h/sintheta = 2/sin35
Wf = -0.2*50*9.8*2/tan35
Wf = -280 J <<<<<<------ANSWER
==========================
d)
normal force is perpendicualr to the motion of body
Wn = Fn*L*cos90
Wn = 0 <<<<<<------ANSWER
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