JEE Mains · Physics · STD 11 - 4.1 newtons laws of motion
A wedge \(Y\) with mass of \(10\) kg and all frictionless surfaces and the inclined surface making \(37°\) with horizontal. A block \(X\) with mass \(2\) kg is placed at the highest point of the wedge as shown in figure is at rest. At \(t = 0\) wedge \((Y)\) is pulled toward right with constant force \((f)\) of \(24\) N. Taking the block \(X\) at rest at \(t = 0\), the time taken by it to slide down \(8.8\) m on the slope, while \(Y\) is on the move, is _____ s.
(take \(\tan(37°) = 3/4\) and \(g = 10\) m/s\(^2\))

- A \(2\)
- B \(4\)
- C \(\sqrt{2}\)
- D \(2\sqrt{2}\)
Answer & Solution
Correct Answer
(A) \(2\)
Step-by-step Solution
Detailed explanation
Horizontal acceleration of the wedge (assuming the block and wedge move together horizontally): \(A = \dfrac{f}{M + m} = \dfrac{24}{10 + 2} = 2\text{ m/s}^2\) Working in the non-inertial frame of the wedge, the block experiences a pseudo force \(mA\) directed opposite to the…
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