Advanced Mathematics - Motion under a constant force - example 1
Motion under a constant force - example 1 continued
b. Equations of motion:
For the larger mass (moving down): 3_g_ -_ T_ = 3_a_
For the smaller mass (moving up): _T_ - 2_g_ = 2_a_
c. After 2 seconds the smaller mass will move up: t = 2, u = 0 (starts
Using the constant acceleration formula _v_ = _u_ + _at_:
d. When the 3kg mass hits the ground after travelling 6m:
When it hits the ground the 3kg mass will be travelling down at 4.85 ms
(assuming that there is plenty of room and the smaller mass does not reach
the pulley first).
Making positive the answers would be:
_v_ = - 3.92 ms for the smaller mass,
_v_ = 4.85 ms for the larger mass.
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