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/s2) [figure]
- 2
- 4
- 2
- 2 2
Answer
(1) 2
Solution
Let the wedge move right with acceleration A and the block slide down the slope with acceleration ar relative to the wedge (m=2 kg, M=10 kg, 37 = 35, 37 = 45).
[figure]
Block in the wedge frame (pseudo force mA towards left):
Perpendicular to the slope: N=m(g 37 +A 37 )=16+1.2A
Along the slope: ar=g 37 -A 37 =6-0.8A
Wedge (horizontal): MA=f-N 37
⇒ 10A=24-0.6(16+1.2A) ⇒ 10.72A=14.4 ⇒ A 1.34 m/s2
∴ ar=6-0.8 1.34 4.9 m/s2
s= 12art2 ⇒ t= 2 8.84.9 1.9 s 2 s
(Taking the whole system's acceleration A= fM+m= 2412=2 m/s2 as an approximation gives ar=6-1.6=4.4 m/s2 and t2= 2 8.84.4=4, i.e. t=2 s exactly.)
∴ t 2 s
Answer: (1)