A block of mass 1 kg is at rest relative to a smooth wedge being moved leftwards with constant acceleration 5m/s2. Let n be the normal reaction between the block and the wedge. Then n and tano are : a = 5 (a) n = 5/5 n and tan 0 = = (b) n = 15 n and tan 0 = - (c) n = 5v5 n and tan 0 = 2 (d) n = 15 n and tan 0 = 2
Answers
Let angle of inclination of wedge is α.
A block of mass m = 1kg is rest relative to the wedge being moved leftward with constant acceleration, a = 5m/s² .
it means, block is placed in non-inertial frame of reference. and we know, Newtonian mechanics is not valid in non-inertial frame of reference.
so, apply Pseudo force on block just opposite direction of motion of wedge.
as block is rest relative to wedge ,
so, horizontal component of Psuedo force , ma = horizontal component of mg
or, mgsinα = macosα
or, tanα = a/g = 5/10 = 1/2
again, Pseudo force = ma is acting horizontally right direction and mg is acting vertically downward direction.
so, net force = √{(ma)² + (mg)²}
= m√(a² + g²)
= 1kg × √(5² + 10²)
= 5√5 N
Answer:
Ok,Answer is 5√5
Explanation:
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