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$$^2$$)

Let’s solve step by step.

step 1: acceleration of wedge

total mass = 10 + 2 = 12 kg

external force = 24 N

$$a_{wedge}=\frac{24}{12}=2m/s^2$$

step 2: go to wedge frame

pseudo force on block (mass 2 kg):

F_p=ma=2×2=4 N(towards left)

step 3: resolve along incline

incline angle = 37°

components:

gravity down the plane:

$$mg\sin⁡37^{\circ\ }=2\times10\times\frac{3}{5}=12N$$

pseudo force component up the plane:

$$4\cos⁡37^{\circ\ }=4\times\frac{4}{5}=3.2N$$

net force down the plane:

12−3.2=8.8 N

step 4: acceleration of block along plane

$$a=\frac{8.8}{2}=4.4m/s^2$$

step 5: motion along incline

distance = 8.8 m
initial velocity = 0

$$s=\frac{1}{2}at^2$$

$$8.8=\frac{1}{2}\times4.4\times t^2=2.2t^2$$

$$2t=4⇒t=2s$$