What Are the Forces Acting on a Box Sliding Down a Ramp?

[mizunopro]

A 115.0 kg box slides down a 25.0 degree ramp with an acceleration of 2.60 m/s squared. Find the normal force, the net force down the ramp, and the force of kinetic friction.

Gravity = 9.81 m/s2
Force of gravity = ma
Mew k = Force of Kinetic Friction/Normal Force

These are the parts that I've been able to find so far...
Normal Force= 1020N --> Force of gravity in the Y direction
The Net Force Down the Ramp= 299N --> 115kg x 2.60
Kinetic Friction = I have 178N and I got this by subtracting 299 from 478, but that was just a guess.

Any ideas?

 

[Kurdt]

Some of the numbers are a little out from what I get, but it looks like you've used the correct method to get them all.

What are you stuck on, or did you just want confirmation?

 

[mizunopro]

i was just looking for some confirmation on the kinetic friction part of the problem. so, anytime a problem asks for kinetic friction and i don't have the mew value, i can just subtract the net force down the ramp from the force of gravity in the x direction? i think i realized that the calculation error was in 478-299..it should be 179 and not 178. thanks for your help..i have a test tomorrow and just wanted to make sure that i had this down.

 

[Kurdt]

i was just looking for some confirmation on the kinetic friction part of the problem. so, anytime a problem asks for kinetic friction and i don't have the mew value, i can just subtract the net force down the ramp from the force of gravity in the x direction? i think i realized that the calculation error was in 478-299..it should be 179 and not 178. thanks for your help..i have a test tomorrow and just wanted to make sure that i had this down.

Yes your methods look good, and if you don't have the coefficient of friction then yes you subtract the net force from whatever force is acting on the object to find the friction force. Do be careful with your numbers and significant figures though, as most people marking a physics test will penalise for that.

 

[mizunopro]

alright. once again, thank you for your help!