Where should I begin with this problem?

  • Thread starter Thread starter jimmyboykun
  • Start date Start date
AI Thread Summary
To solve the problem of work done on a 150-kg crate pushed across a wooden floor, the work done by the person pushing the crate is calculated using the formula W=F*s, where F is the force and s is the distance. The frictional force, determined using the kinetic coefficient of friction, is 441.45N. The work done by the person is then calculated as 1986.525J, but the answer should be rounded to two significant figures, resulting in 2.0 kJ. The work done by friction is negative, equal to the calculated value. The incorrect significant figures likely contributed to the low score received on the problem.
jimmyboykun
Messages
39
Reaction score
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

 
Physics news on Phys.org
Show your work.
 
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?
 
jimmyboykun said:
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.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top