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A block moves up an inclined plane as follow and later returns to its starting position.
Why the F-t graph of the block for the whole journey is as follow? Taking the direction moving up the plane as positive. Assume the magnitude of friction acting on the block is unchanged.
- 天同Lv 78 年前最愛解答
In the upward journey, the net force acting on the block Fu is,
Fu = -(mg.sin(a) + Ff)
where m is the mass of the block
g is the acceleration due to gravity
a is the angle of the inclined plane
Ff is the frictional force.
Because both mg.sin(a) and Ff are pointing downward parallel to the slope, Fu is -ve. Hence, the F-t graph is a straight line, with ordinate (i.e. value on y-axis) equals to -(mg.sin(a) + Ff), parallel to the time axis (t-axis)
When the block moves downward, the net force Fd is,
Fd = -mg.sin(a) + Ff
Now, the frictional force, which is always in opposite direction to the motion of the block, points upward along the slope.
Hence, Fd = -(mg.sin(a) - Ff)
Thus, Fd < Fu in magnitude.
The graph will be a straight line parallel to the t-axis, but with absolute value less than that in the upward journey.
The "step" on the graph is the time when the block reaches the highest position on the slope.