Moderators
______________
______________
Ques.=> Starting from rest, a body slides down at 45° inclined plane in twice the time it takes to slide down the same distance in the absence of friction. The coefficient of friction between the body and the inclined plane is???
______________________
Class => 11th
Sub. => physics
______________________
Jai bajarang bali
@DrSM
Answers
Hy mate..
______________________
______________________
Your question is
______________________
Q. Starting from rest, a body slides down at 45° inclined plane in twice the time it takes to slide down the same distance in the absence of friction. The coefficient of friction between the body and the inclined plane is?
______________________
HERE IS YOUR ANSWER..
______________________
STEP 1.
✔️The normal force, N = mgsin45°
✔️ The friction force, f = µN
The weight force W = mgcos45°
(Not weight of object but weight force
____________________________
along incline.
____________
STEP 2.
➡️ The net force F = W-f
➡️ Let F1 = Force with no friction thus.
➡️ F1 = W
➡️ F2 = Force with friction, thus
F2 = W-f
➡️ a1 = F1/m
➡️ F2/m
s = Vot + 1/2at^2
____________
Assume Vo = 0
(although it subtracts out anyway if it
is equal to something)
Since s is the same for both situations. and
t2 = 2×t1,
STEP 3.
➡️ 1 /2 (a1) (t1)^2
➡️ 1 /2 (a2) (2×t1)^2
➡️ (a1)/(a2) = 4
➡️ F1/F2 = 4
➡️ W/(W-f) = 4
W-f = W/4
➡️ (3/4)W = f
.75(mgcos45°) = µ(mgsin45°)
➡️ µ = .75
THANKYOU
______________
BRAINLY USER