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Question

# In figure, a crate slides down an inclined right angled trough. Acceleration of the crate is

A
2.1 m/s2
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B
2.5 m/s2
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C
2.7 m/s2
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D
2.9 m/s2
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Solution

## The correct option is A 2.1 m/s2FBD of crate: Here, N= Normal reaction on the crate applied by trough, f= Friction force on the crate opposite to the sliding motion. Balancing forces for front view of the crate, we have, mgcos45∘=√N2+N2 ⇒√2N=mg√2 ⇒N=mg2 Now, balancing forces for side view, we have, mgsin45∘−2f=ma......(1) where, a= acceleration of the crate parallel to the inclined plane in the downwards direction. f=μN=0.5N=0.5mg2 Substituting the values in (1), we have, mgsin45∘−2μN=ma ⇒ma=mg×1√2−2×0.5×mg2 ⇒a=g√2−0.5g ⇒a=0.71−0.5g ⇒a=0.21g ⇒a=0.21×10 ∴a=2.1 m/s2 Hence, option (a) is correct answer.

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