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Answer:
The magnitude of the acceleration of the 19 kg block is 1.414 m/s²
Explanation:
From Newton's second law of motion;
where;
m is the mass of the objects involved, kg
a is the acceleration of the object, m/s²
different forces on the block and string
⇒force due to 15 kg block
=mgcosθ = 15×9.8×cos20 = 15×9.8×0.9396
= 138.12 N
⇒Tensional Force on 19 kg mass:
T = mg = 19×9.8 = 186.2 N
186.2 - 138.12 = a(15+19)
48.08 = a(34)
a = 48.08/34
a = 1.414 m/s²
Therefore, the magnitude of the acceleration of the 19 kg block is 1.414 m/s²
Explanation:
Given that,
Mass of the block, m = 12.2 kg
Initial velocity of the block, u = 6.65 m/s
The coefficient of kinetic friction,
(a)The force of kinetic friction is given by :
mg is the normal force
So,
(b) Net force acting on the block in the horizontal direction,
f = ma
a is the acceleration of the block
(c) Let d is the distance covered by the block before coming to the rest. Using third equation of motion as follows :
Hence, this is the required solution.