Laws Of Motion Question 244
Question: A block of mass m is placed on a smooth wedge of inclination $ \theta $ . The whole system is accelerated horizontally so that the block does not slip on the wedge. The force exerted by the wedge on the block (g is acceleration due to gravity) will be [CBSE PMT 2004]
Options:
A) $ mg\cos \theta $
B) $ mg\sin \theta $
C) $ mg $
D) $ mg/\cos \theta $
Show Answer
Answer:
Correct Answer: D
Solution:
When the whole system is accelerated towards left then pseudo force (ma) works on a block towards right.
For the condition of equilibrium $ mg\sin \theta =ma\cos \theta $
therefore $ a=\frac{g\sin \theta }{\cos \theta } $
$ \therefore $ Force exerted by the wedge on the block $ R=mg\cos \theta +ma\sin \theta $ R$ =mg\cos \theta +m( \frac{g\sin \theta }{\cos \theta } )\sin \theta $
$ =\frac{mg({{\cos }^{2}}\theta +{{\sin }^{2}}\theta )}{\cos \theta } $ R$ =\frac{mg}{\cos \theta } $
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