CBSE Class 10 Answered
A body is projected up a 40 degrees rough inclined plane. If the coefficient friction is 0.5, then find the retardation of block.
Asked by acv27joy | 20 May, 2018, 09:02: PM
Expert Answer
frictional force acting on the block in retardation
f = uN (N is normal force)
N = mgcosθ
f = umgcosθ
component of weight acts in backward direction
Fb = mgsinθ
total force in downward direction is F+f
FT = umgcosθ + mgsinθ
maT= mg(ucosθ+sinθ)
aT= g(ucosθ+sinθ)
Answered by Science Mate | 21 May, 2018, 01:19: AM
Application Videos
Concept Videos
CBSE 10 - Physics
Asked by agankitgupta938 | 18 Apr, 2024, 04:29: PM
ANSWERED BY EXPERT
CBSE 10 - Physics
Asked by infinityupgraded | 13 Apr, 2024, 08:17: AM
ANSWERED BY EXPERT
CBSE 10 - Physics
Asked by suryamr2019 | 08 Mar, 2024, 04:32: PM
ANSWERED BY EXPERT
CBSE 10 - Physics
Asked by sheetal.kolte | 04 Mar, 2024, 12:38: PM
ANSWERED BY EXPERT
CBSE 10 - Physics
Asked by shrilakshmimunoli | 01 Mar, 2024, 01:15: AM
ANSWERED BY EXPERT
CBSE 10 - Physics
Asked by khajannirwan | 27 Feb, 2024, 10:20: PM
ANSWERED BY EXPERT
CBSE 10 - Physics
Asked by sailakshmi.avinesh | 13 Feb, 2024, 07:03: AM
ANSWERED BY EXPERT
CBSE 10 - Physics
Asked by saurabhjd527 | 30 Jan, 2024, 07:55: PM
ANSWERED BY EXPERT
CBSE 10 - Physics
Asked by saanviyadla | 24 Jan, 2024, 07:06: PM
ANSWERED BY EXPERT