here,
A)
mass ,m = 1.6 kg
the net force , F = 5 N i + 2.82 j N - 1 N i + 3 N * ( - sin(20) i - cos(20) j)
F = 2.97 i N
the accelration , a = F/m
a = 2.97 /1.6 m/s^2 = 1.86 m/s^2 i
the x-component of the accelration is 1.86 m/s^2
B)
the y-component of the accelration is 0 m/s^2
C)
the net force , F = - 3 N i + 1.414 j N + 2 * ( cos(15)i + sin(15) j) + 2 N * ( sin(15) i - cos(15) j)
F = - 0.55 i N + 0 j N
the accelration , a = F/m
a = 0.55 /1.6 m/s^2 = - 0.34 m/s^2 i
the x-component of the accelration is - 0.34 m/s^2
d)
the y-component of the accelration is 0 m/s^2
Part A In each of the two free-body diagrams, the forces are acting on a 1.6...
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