Answers:
a) 385.74 J
b) -98.1 J
c) 192.87 N
Explanation:
A figure is attached in order to understand better the question. In addition, we have the following data:
is the mass of the block
is the distance the block is pushed at constant velocity
is the angle at wich the force
is applied
is the acceleration due gravity
is the coefficient of kinetic friction between the block and the wall
Now, if we draw a free body diagram of this situation we will have the following:
![\sum{F_{netx}}=-N+F_{x}=0](https://tex.z-dn.net/?f=%5Csum%7BF_%7Bnetx%7D%7D%3D-N%2BF_%7Bx%7D%3D0)
Where
is the Normal force and
is the horizontal component of the applied force ![F](https://tex.z-dn.net/?f=F)
Then:
(1)
![\sum{F_{nety}}=F_{y}-F_{friction}-Fg=0](https://tex.z-dn.net/?f=%5Csum%7BF_%7Bnety%7D%7D%3DF_%7By%7D-F_%7Bfriction%7D-Fg%3D0)
Where
is the vertical component of the applied force
,
is the Friction force and
is the force due gravity (the weight of the block)
Then:
(2)
Finding
from (2):
(3)
(4)
(5)
<h3>a) Now we can calculate <u>the work done by the force
![F](https://tex.z-dn.net/?f=F)
:</u></h3><h3 />
When the applied force and the direction of motion form an angle the expression to calculate the Work
is:
(6)
(7)
(8) This is the work done by ![F](https://tex.z-dn.net/?f=F)
<h3>
b) The work done by gravity on the block</h3>
In this case the work equation is:
(9)
Where:
is the gravity force on the block
is the angle between the gravity force and the direction of motion
(10)
(11) Note the work is negative because the applied force is in the opposite direction of motion
<h3>
c) The magnitude of the normal force</h3>
From (1) we know the Normal force is:
(1)
Since
, we have:
(12)
(13) This is the Normal force