Q 7.127:
A single force acts on a 4.3 kg4.3 kg particle-like object in such a way that the position of the object as a function of time is given by x = 4.2t + 3.1t2 + 2.2t3x = 4.2t + 3.1t2 + 2.2t3 , with xx in meters and tt in seconds. Find the work done on the object by the force from t = 1.4 st = 1.4 s to t = 3.5 st = 3.5 s .
The answer is 23067
given
mass m = 4.3 kg
x = 4.2 t + 3.1 t2 + 2.2 t3
t = 1.4 sec to 3.5 sec
v = dx/dt = d ( 4.2 t + 3.1 t2 + 2.2 t3 ) / dt
v = 4.2 + 6.2 t + 6.6 t2
v1.4 = 4.2 + 6.2 X 1.4 + 6.6 X 1.42
v1.4 = 25.81 m/sec
sameway
v = dx/dt = d ( 4.2 t + 3.1 t2 + 2.2 t3 ) / dt
v = 4.2 + 6.2 t + 6.6 t2
v3.5 = 4.2 + 6.2 X 3.5 + 6.6 X 3.52
v3.5 = 106.75 m/sec
W = 1/2 m ( V3.52 - V1.42 )
W = 0.5 X 4.3 X ( 106.752 - 25.812 )
W = 2.15 X ( 10729.4064 )
W = 23068.22 J
work done on the object by the force from t = 1.4 s to t = 3.5 s is W = 23068.22 J
Q 7.127: A single force acts on a 4.3 kg4.3 kg particle-like object in such a...
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