A wedge $Y$ with mass of 10 kg and all frictionless surfaces and the inclined surface making $37^\circ$ 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 $\tan (37^\circ) = 3/4$ and $g = 10 \text{ m/s}^2$)
Step 1: Understanding the Question:
We have a block on an accelerating wedge. We need to find the time it takes for the block to slide a certain distance along the wedge's surface. This is a problem of relative motion in a non-inertial frame. Step 2: Key Formula or Approach:
1. Find Wedge Acceleration: Calculate the horizontal acceleration of the combined wedge-block system, as there's no friction between them.
2. Analyze forces in Wedge's Frame: Analyze the motion of the block from the perspective of the accelerating wedge. This requires introducing a pseudo-force ($F_{pseudo} = -m_{block}A_{wedge}$) acting on the block.
3. Find Relative Acceleration: Sum the forces acting on the block parallel to the incline (gravity component and pseudo-force component) to find its net acceleration relative to the wedge.
4. Use Kinematics: Apply the equation of motion $s = ut + \frac{1}{2}at^2$ to find the time. Step 3: Detailed Explanation: Part 1: Acceleration of the Wedge
The external horizontal force $f=24$ N acts on the entire system (wedge Y + block X).
Total mass $M_{total} = M_Y + M_X = 10 + 2 = 12$ kg.
The horizontal acceleration of the wedge (and the block initially) is:
\[ A = \frac{f}{M_{total}} = \frac{24 \text{ N}}{12 \text{ kg}} = 2 \text{ m/s}^2 \text{ (to the right)} \]
Part 2: Relative Acceleration of the Block
Let's analyze the forces on block X in the frame of reference of the wedge.
- The component of gravity acting down the incline is $mg \sin(37^\circ)$.
- Since the wedge accelerates to the right, a pseudo-force $F_{pseudo} = mA$ acts on the block to the left.
- The component of this pseudo-force acting up the incline is $(mA) \cos(37^\circ)$.
The net force on the block along the incline (taking down as positive) is:
$F_{net, rel} = mg \sin(37^\circ) - mA \cos(37^\circ)$.
The relative acceleration $a_{rel}$ is $F_{net, rel}/m$:
\[ a_{rel} = g \sin(37^\circ) - A \cos(37^\circ) \]
Given $\tan(37^\circ) = 3/4$, we have $\sin(37^\circ) = 3/5$ and $\cos(37^\circ) = 4/5$.
\[ a_{rel} = 10\left(\frac{3}{5}\right) - 2\left(\frac{4}{5}\right) = 6 - \frac{8}{5} = 6 - 1.6 = 4.4 \text{ m/s}^2 \]
Part 3: Calculate the Time
The block starts from rest ($u=0$) and slides a distance $s = 8.8$ m down the incline with acceleration $a_{rel} = 4.4 \text{ m/s}^2$.
Using the kinematic equation $s = ut + \frac{1}{2}at^2$:
\[ 8.8 = 0 \cdot t + \frac{1}{2}(4.4)t^2 \]
\[ 8.8 = 2.2 t^2 \]
\[ t^2 = \frac{8.8}{2.2} = 4 \]
\[ t = \sqrt{4} = 2 \text{ s} \]
Step 4: Final Answer:
The time taken by the block to slide down is 2 s.
