A block of mass 10 kg is moving up an inclined plane of inclination 30° with an initial speed of 5 m/s. It stops after 0.5 s, what is the value of coefficient of kinetic friction?
Text Solution
Verified by ExpertsD
For equilibrium, normal to pane
N = mg cos θ …
Net force along the plane downward
F = mg sin θ + fk …
where fk is kinetic friction
but fk = μ N = μ mg cos θ …
from eq. , , and we get
∴ F = mg sin θ + μ mg cos θ
According to Newton’s IInd law
F = ma
∴ ma = mg sin θ + μ mg cos θ
∴ Retardation a = g sin θ + μ g cos θ
From equation v = u + at, (we have)
o = u – (g sin θ + μ g cos θ ) t
⇒ g sin θ + m g cos θ = 
⇒ 10 × sin 30° + × 10 cos 30° = 
⇒ 10 × 
+ 10 μ × 
= 10
⇒ 5 
μ = 5
or μ = 
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems