Work and energy and that sort of rot. :/

AI Thread Summary
Sally applies a horizontal force of 520 N at an angle of 53.0° to drag a wooden crate across the floor at a constant speed. The force exerted by the rope on the crate is calculated to be 864.05 N, and the work done by Sally in moving the crate 28.5 m is 14,820 J. However, there is uncertainty regarding how to calculate the work done by the floor through friction, as the relationship between friction force, normal force, and the coefficient of friction needs clarification. The discussion emphasizes the importance of understanding how the angle of the applied force affects the normal force. Overall, the calculations for the first two parts are correct, but further assistance is needed for the friction component.
DDRchick
Messages
27
Reaction score
0
Sally applies a horizontal force of 520 N with a rope to drag wooden crate across a floor with a constant speed. The rope tied to the crate is pulled at an angle of 53.0°.

(a) How much force is exerted by the rope on the crate?
Solved and correct.

(b) What work is done by Sally if the crate is moved 28.5 m?
Solved and correct

(c) What work is done by the floor through force of friction between the floor and the crate?
This one I have no idea how to do, unfortunately :(
FxD=Work
Trig (cos, sine, tan)
Work=F(d)+(friction force)(d)
work= change in Kinetic Energy
Potential Energy=mass(g)(Height)


(a) How much force is exerted by the rope on the crate?
I managed to figure this one out.
cos53=520N/x
cos53x=520
x=864.05 N

(b) What work is done by Sally if the crate is moved 28.5 m?
All I did with this one was multiply force (520) by distance (28.5) and got 14820 J

(c) What work is done by the floor through force of friction between the floor and the crate?
This one I have no idea how to do, unfortunately :(


I've unfortunately procrastinated and it's due tomorrow but I'm sure I'll manage if someone doesn't reply...:/
 
Physics news on Phys.org
(c) What work is done by the floor through force of friction between the floor and the crate?
This one I have no idea how to do, unfortunately :(

Work is Force X Distance, right? What is the Force of friction? How is it defined in terms of the normal Force on the floor, and the coefficient of (dynamic, sliding) friction?
 
berkeman said:
Work is Force X Distance, right? What is the Force of friction? How is it defined in terms of the normal Force on the floor, and the coefficient of (dynamic, sliding) friction?

Also keep in mind that the Fn isn't going to just equal the Fw because it is being pulled at an angle and some of that force is going into the y-component.
 
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