How do elastic objects behave when different forces are applied?
1 answer:
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Answer:
v = 79.2 m/s
Solution:
As per the question:
Mass of the object, m = 250 g = 0.250 kg
Angle,
Coefficient of kinetic friction,
Mass attached to the string, m = 0.200 kg
Distance, d = 30 cm = 0.03 m
Now,
The tension in the string is given by:
(1)
Also
T = m(g + a)
Thus eqn (1) can be written as:
Now, the speed is given by the third eqn of motion with initial velocity being zero:
where
u = initial velocity = 0
Thus