Where to start with this problem?

  • #1
jimmyboykun
39
0

Homework Statement



You are pushing a 150-kg wooden crate in a straight line a distance of 4.5 m across a wooden floor at constant speed. The static and kinetic coefficients of friction are 0.42 and 0.30, respectively. What is the work done by you on the crate? and What is the work done by friction on the crate?

I understand that with constant speed meas that my acceleration is zero, but after that I am lost.

Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
NihalSh
199
15
Show your work.
 
  • #3
jimmyboykun
39
0
ok here's what I did work is W=F*s.

I multiply 0.30(kinetic energy) because the crate is in motion by 150kg by 9.81m/s^2 which gave me 441.45N as my frictional force. Now I have my force I will multiply it by my displacement to give me my work.

441.45N*4.5m= 1986.525J.

I did it and submitted the answer and received a 1/10 from this problem. What did I do wrong?
 
  • #4
NihalSh
199
15
ok here's what I did work is W=F*s.

I multiply 0.30(kinetic energy) because the crate is in motion by 150kg by 9.81m/s^2 which gave me 441.45N as my frictional force. Now I have my force I will multiply it by my displacement to give me my work.

441.45N*4.5m= 1986.525J.

I did it and submitted the answer and received a 1/10 from this problem. What did I do ##wrong?
Your solution seems fine for "work done by you". "work done by friction" would be negative of the calculated value. Also note that your answer should have only two significant figures. That means answer should have been ##2.0~kJ##. This could be the reason you were marked down.
 

Suggested for: Where to start with this problem?

Replies
11
Views
431
  • Last Post
Replies
2
Views
259
Replies
1
Views
281
Replies
5
Views
321
Replies
15
Views
338
  • Last Post
Replies
1
Views
166
Replies
7
Views
338
  • Last Post
2
Replies
44
Views
647
Replies
3
Views
457
Replies
13
Views
421
Top