A 400 g particle moving along the z-axis experiences the force shown in (Figure 1). The particle's velocity is 2.0m/s at x = 0m. You may want to review (Pages 214 - 218)
Part A
What is its velocity at x= 3 m?
Express your answer to two significant figures and include the appropriate units.
ANSWER :
Mass of particle = 400 g = 0.4 kg
Force applied = 15 N from x = 0 to x = 1.
Now,
F = m * a
=> a = F/m = 15/0.4 = 37.5 m/sec^2 .
Let t be the time taken to move from x = 0 to x = 1.
So,
As per formula : s = u t + 1/2 a t^2
=> 1 = 2.0 t + 1/2 (37.5) * t^2
=> 18.75 t^2 + 2.0 t - 1 = 0
=> t = - 2/ (2*18.75) +/- sqrt(2^2 - 4*18.75*(-1)) / (2*18.75)
=> t = - 0.29 , 0.1837 secs.
Discard the negative value which is meaningless.
=> t = 0.1837 sec.
Hence, velocity at x = 1, V1 = u + a t = 2 + 37.5 * 0.1837 = 8.89 m/sec
From x = 1 to x = 3, F is decreasing at a constant rate. So, we can take average F as acting during this interval.
Hence, F’ = (15 + 0)/2 = 7.5 N in the interval (1, 3).
So, acceleration a’ in that interval = F’/a’ = 7.5/0.4 = 18.75 m/sec^2
For moving from x = 1 to x = 3, if time taken is t’,
then :
s = V1 t’ + 1/2 a’ t’^2
=> 2 = 8.89 * t’ + 1/2 * 18.75 t’^2
=> 9.375 t’^2 + 8.89 t’ - 2 = 0
=> t’ = - 8.89/(2*9.375) + sqrt(8.89^2 - 4*9.375*(-2)) / (2*9.375)
(We have discarded the negative value of t’)
=> t ‘ = 0.1878 sec.
So final velocity at x = 3, V3
= V1 + a’ t’
=> V3 = 8.89 + 18.75 * 0.1878
=> V3 = 12.41 m/sec.
So, velocity at x = 3, is 12.41 m/sec (ANSWER).
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