A magician pulls a tablecloth from under a 200 g mug located 30.0 cm from the edge of the cloth. The cloth exerts a friction force of 0.105 N on the mug, and the cloth is pulled with a constant acceleration of 3.50 m/s2. How far does the mug move relative to the horizontal tabletop before the cloth is completely out from under it? Note that the cloth must move more than 30 cm relative to the tabletop during the process.
consider the mug: F=ma you know F, you know the time of force (time=distance/velocitycloth), so
Force*time=massmug*velocitymug
from that,you caculate the averagevelocity of the mug (1/2 velocityfinalmug), then distance=avgvelocity*time
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To find how far the mug moves relative to the tabletop before the cloth is completely out from under it, we can use the equation of motion:
s = ut + (1/2)at^2
Where:
s = displacement
u = initial velocity (0 m/s since the cloth is pulled from rest)
a = acceleration (3.50 m/s^2)
t = time
First, let's find the time it takes for the cloth to completely move out from under the mug.
We know that the distance the mug is from the edge of the cloth is 30.0 cm = 0.30 m.
Since the mug is pulled with a constant acceleration, we can use the equation to find the time it takes for the cloth to move a distance of 0.30 m:
s = ut + (1/2)at^2
0.30 = 0(t) + (1/2)(3.50)(t^2)
0.30 = 1.75t^2
t^2 = 0.30 / 1.75
t^2 = 0.1714
t ≈ √(0.1714)
t ≈ 0.4147 seconds
Now that we have the time, we can find the displacement of the mug during this time.
Using the equation of motion:
s = ut + (1/2)at^2
s = 0(0.4147) + (1/2)(3.50)(0.4147^2)
s ≈ (1/2)(3.50)(0.1719)
s ≈ 0.3009 m
Therefore, the mug moves approximately 0.3009 meters relative to the horizontal tabletop before the cloth is completely out from under it.