Khai Lim

I try to looking for this answer but now I think I am a person should answer this question ^^. Ok Sum F = ma ( because it get loading the crates ) Fp - F(fr) = ma Fp = ma + F(fr) Fp = ma + u(k) mg => u(k) = (Fp - ma) / mg ( with ma is 0 ) so u(k) = 59/ (...

I try to looking for this answer but now I think I am a person should answer this question ^^. Ok Sum F = ma ( because it get loading the crates ) Fp - F(fr) = ma Fp = ma + F(fr) Fp = ma + u(k) mg => u(k) = (Fp - ma) / mg ( with ma is 0 ) so u(k) = 57/ (...

W = (Fcos(theta))*vt

The process is correct.

KE = PE + W_friction 1/2 * mv^2 = mgh + mu*mgcos(theta)d ----- m is cancel out --- sin(theta) = heigh/ distance( inclide) ===> h = d*sin(theta) 1/2 v^2 = g (d*sin(theta)) + mu* (g*cos(theta))*d now you can solve for mu ^^ Good luck.

KE = PE + W_friction 1/2 * mv^2 = mgh + mu*mgcos(theta)d ----- m is cancel out --- sin(theta) = heigh/ distance( inclide) ===> h = d*sin(theta) 1/2 v^2 = g (d*sin(theta)) + mu* (g*cos(theta))*d now you can solve for mu ^^ Good luck.