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2 answers:
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Answer:
a, b) part a and b on diagram attached
c) sf = -35 i + 20 j
35 km West and 20 km North
Explanation:
For part a and b refer to the attached co-ordinate system:
Note: unit vector i is in West/East direction and unit vector j is in North/South direction.
si = -15 i + 25 j
sf-si = -20 i - 5 j
Hence,
Mark relative position from habitat sf = si + sf/i
sf = ( -15 i + 25 j ) + ( -20 i - 5 j )
sf = -35 i + 20 j
35 km West and 20 km North
Answer:
A) a frictional force is acting on it
Explanation:
The crate is sliding down the inclined ramp at constant velocity: constant velocity means zero acceleration, a = 0. According to Newton's second law:
F = ma
this also means that the net force acting along the direction of the slope is zero.
Without frictional force, there would be only one force acting along this direction: the component of the weight of the crane parallel to the slope, acting downward:
where is the angle of the incline. Therefore, the net force along this direction would not be zero. In order to have a net force of zero, there must be another force acting upward on the crate: and the only force that is possibly acting on the crate against its direction of motion is the frictional force, whose magnitude must be equal to
, in order to produce a net force of zero (and therefore, a zero acceleration).