× Didn't find what you were looking for? Ask a question
Top Posters
Since Sunday
8
5
z
4
n
4
t
4
k
3
x
3
r
3
m
3
j
3
c
3
l
3
Home Q & A Board Science-Related Homework Help Physics
New Topic  
datadisha datadisha
wrote...
Posts: 19
Rep: 0 0
11 years ago
A 330-kg piano slides 3.6 m down a 28 dagree incline and is kept from accelerating by a man who is pushing back on it parrallel to the incline.  The effective coefficient of kinetic friction is 0.40. How do you calculate the work done by the friction foce,and the work done by the force of gravity?
Read 633 times
2 Replies

Related Topics

Replies
wrote...
#1
11 years ago
Force normal to the plane is mg*cos(28) = 2855.5 newtons
force of friction against movement = 0.40* 2855.5 = 1142.2 N

work by friction = force of friction * distance = 1142.2 N* 3.6 m
4111.9 newton meters

summation of forces parallel to the plane:

0 = mg*sin(28) -4111.9 N -man's push

Still, mg*sin(28)* distance is the work done by gravity 5465.8 J
wrote...
#2
11 years ago
a)Wf= Ff*S=(N*f)*S
Wf=(m*g*cos28*f)*S
Wf=(330*10*cos28*0.4)*3.6
Wf=4196 Joules
b)Wg=Fg*S
Wg=(m*g*sin28)*S
Wg=(330*10*sin28)*3.6
Wg=5577 Joules
Wt=Wf+Wg
Wt=4196+5577
Wt=9773 Joules
New Topic      
Quick Reply
Explore
Post your homework questions and get free online help from our incredible volunteers
  911 People Browsing
 111 Signed Up Today
Start New Topic Take the Tour CoursesNew Study Tools Topics Trending Browse by Textbook
Live Chat
Related Images
  
 857
  
 5694
  
 181
Your Opinion
Which 'study break' activity do you find most distracting?
Votes: 824
Previous poll results: Who will win the 2024 president election?