As shown in the figure below, a box of mass m = 57.0 kg (initially at rest) is p
ID: 1338609 • Letter: A
Question
As shown in the figure below, a box of mass
m = 57.0 kg
(initially at rest) is pushed a distance
d = 79.0 m
across a rough warehouse floor by an applied force of
FA = 242 N
directed at an angle of 30.0° below the horizontal. The coefficient of kinetic friction between the floor and the box is 0.100. Determine the following. (For parts (a) through (d), give your answer to the nearest multiple of 10.)
(d) work done by the force of friction
___ J
(e) Calculate the net work on the box by finding the sum of all the works done by each individual force.
WNet = ___ J
WNet = ____ J
Explanation / Answer
(D)
Work done by friction force is:
W(f) = F(f) x d, where F(f) is the Frictional force and d is the distance moved, and F(f) = ?N, where ? is the coefficient of friction
In this case, we're talking about a piano of mass, m = 57 kg, sliding down 79m on an incline that's at 30 degrees. Therefore:
Here the work done by Friction is:
Wf= Ff*S=(N*f)*S
Wf=(m*g*cos28*f)*S
wf=(57*10*cos30*0.100)*79
Wf=3899.71joules
(e) work done by gravity
Wg=Fg*S
Wg=(m*g*sin30)*S
Wg=(57*10*0.5)*79
Wg=22515joules
total workdone is
Wt=Wf+Wg
Wt=3899.71+22515
Wt=26414.71
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