Answer:
(a) T= 38.4 N
(b) m= 26.67 kg
Explanation:
We apply Newton's second law:
∑F = m*a (Formula 1)
∑F : algebraic sum of the forces in Newton (N)
m : mass in kilograms (kg)
a : acceleration in meters over second square (m/s²)
Kinematics
d= v₀t+ (1/2)*a*t² (Formula 2)
d:displacement in meters (m)
t : time in seconds (s)
v₀: initial speed in m/s
vf: final speed in m/s
a: acceleration in m/s²
v₀=0, d=18 m , t=5 s
We apply the formula 2 to calculate the accelerations of the blocks:
d= v₀t+ (1/2)*a*t²
18= 0+ (1/2)*a*(5)²
a= (2*18) / ( 25) = 1.44 m/s² to the right
We apply Newton's second law to the block A
∑Fx = m*ax
60-T = 15*1.44
60 - 15*1.44 = T
T = 38.4 N
We apply Newton's second law to the block B
∑Fx = m*ax
T = m*ax
38.4 = m*1.44
m= (38.4) / (1.44)
m = 26.67 kg
The tension on the rope due to motion of the blocks when the force is applied is 21.6 N.
The mass of the block B is 26.67 kg.
The given parameters;
The acceleration of the blocks is calculated as follows;
s = ut + ¹/₂at²
18 = 0 + ¹/₂(5)²a
18 = 12.5a
[tex]a = \frac{18}{12.5} \\\\a = 1.44 \ m/s^2[/tex]
The tension on the rope due to motion of the blocks when the force is applied is given as;
T = ma
T = 15 x 1.44
T = 21.6 N
The net force on block is calculated as follows;
F - T = ma
60 - 21.6 = ma
38.4 = ma
38.4 = 1.44m
[tex]m = \frac{38.4}{1.44} \\\\m = 26.67 \ kg[/tex]
Thus, the mass of the block B is 26.67 kg
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