Starting from rest. you push a 5 kg box with force 20 N across a rough floor (mu
ID: 1704582 • Letter: S
Question
Starting from rest. you push a 5 kg box with force 20 N across a rough floor (mu = 0.3) for 6 m. Compute The acceleration of the box to two decimal places. (Note: * starred answers at bottom of page 3.) The final velocity to two decimal places Now- let's use the new physics. Compute The final kinetic energy, using your answer to (b). The work done by you on the box. That's work done by your push: don't include friction. The force of friction. The work done on the box by friction. Note: sign matters lt's possible to do negative work. The net work done on the box. You're pushing on level ground, with no friction heat being produced. so all the work goes into kinetic energy. Does your answer to (g) match your answer to (c)? This a why we use KE = (1/2)m/v. Answers to starred questionsExplanation / Answer
m=5kg
F=20N (constant)
=0.3
d=6m
v0=0
(a)a=? |( b)vf=? | (c)Kf=? | (d)Wby force=? | (e)Ff=? | (f)Wby frict=? | (g)Wnet=?
a)
F+Ff=ma
a=F+Ff/m where Ffriction=*Fnormal=*m*g=|0.3*5kg*9.81m/s2|=|14.715kgm/s2|=-14.715N
Ffriction=-14.715N -- you will also need this for later
a=20N+(-14.715N)/5kg
a=1.06m/s2
b)
vf2=v02+2*a*d
vf2=0+2*1.06m/s2*6m
vf2=12.72m2/s2
vf=sqrt(12.72m2/s2)
vf=3.57m/s
c)
Kf=mvf2/2
Kf=5kg*12.72m2/s2 / 2
Kf=31.8kgm2/s2
d)
W=F*d
Wby force=20N*6m
Wby force=120Nm=120J
e)
I already found this when solving part (a)
Ff=|*Fnormal| we put it as an absolute value
Ff=|*m*g|
Ff=|0.3*5kg*9.81m/s2|
Ff=|14.72kgm/s2| but (kgm/s2) is actually (Newtion) so
Ff=|14.72N|
Ff=-14.715N because it opposes the applied force it is Negative
f)
Wby frict=Ff*d
Wby frict=-14.715N*6m
Wby frict=-88.3Nm=-88.3J
The answer THEY have given you is WRONG by a (0.1). They did not round correctly.
g)
Wnet=Wby force + Wby frict
Wnet=120Nm+(-88.3Nm)
Wnet=31.7Nm
They got 31.8 because they used the WRONG -88.2 for the work by friction.
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