A) constant acceleration\[~=g\text{ }sin\text{ }\theta \]
B) constant acceleration\[=(g\,\sin \,\theta -\mu \,g\,\cos \theta )\]
C) constant retardation \[=(\mu \,g\,\,\cos \,\theta -g\sin \theta )\]
D) variable acceleration that first decreases from \[g\text{ }sin\text{ }\theta \] to zero and then becomes negative.
Correct Answer: D
Solution :
Force of friction is \[F=\mu R=\mu \,mg\cos \theta \] |
When the body slides down, the downward force along the plane = component \[mg\sin \theta \]of the weight mg. since the force of friction acts upwards along the plane, the effective downward force = \[mg\sin \theta -\mu mg\cos \theta =mg(sin\theta -\mu cos\theta )\] |
\[\therefore \]Acceleration = force/ mass = \[g(sin\theta -\mu cos\theta )\] |
\[=g(\sin \theta -kx\cos \theta )\] |
Hence, the acceleration varies with x and decreases as x increases. Thus, the correct choice is [D]. |
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