a crate of 50 kg slide donw 30 incline, crates acceleration is 2.0m/s^2 and incl

Question

a crate of 50 kg slide donw 30 incline, crates acceleration is 2.0m/s^2 and incline is 10m long

a) what is ke at bottom?

b) how much work is spent in overcoming friction

c) what is magnitude of Friction force act on crate as it slide down incline.

for part A, can't i use the C.O.E ? (conservation of energy) Ei=Ef instead of useing kinematic..

because i wanna try with c.o.e but answer does not come as right.. the answer should be 1000j

let me show my work for part A.

ke=pe

1/2mvf^2 +mghf (zero) = 1/2mvi^2(zero) +mghi

left with, 1/2mvf^2=mghi

so, after solving for vf, which comes out to be 9.8m/s i plug back into equation "1/2mvf^2=mghi "

so, 1/2(50kg)(9.8m/s)=(50kg)(9.8)(sin30*10). h=sin30*10m

then, it comes out as 2401 N= 2450 N ???? iam suppose to get 1000j..... i dont get it help from here.

please dont use kinematic .. i wanna try c.o.e

Explanation / Answer

You can't use COE in this case as the acceleration of the crate is given to be 2 m/s^2 which is not equal to gsin(30). This means that there is some external force, forex friction which should also come into the energy conservation equation. Hence, we use kinematics to calculate the final velocity of the crate at the bottom.

Thanks!

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