Newton's Law's - Calculate time when given distance and tension

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The discussion centers on calculating the time taken for a wooden block, initially at rest, to reach the edge of a table when pulled by a string with a tension of 0.2 mN. Using Newton's Second Law (F = ma), the acceleration of the block can be determined as 0.2 m/s², assuming the mass is represented as a dimensionless variable. The relevant kinematic equation to find the time is s = ut + 1/2 at², where s is the distance (1.6 m), u is the initial velocity (0 m/s), and a is the acceleration (0.2 m/s²).

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Nicaragua
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1. A wooden block of mass m kg is at rest on a table, 1.6 metres from the edge. The block is pulled directly towards the edge by a horizontal string. The tension in the string has magnitude 0.2 mN. Calculate the time taken for the block to reach the edge of the table.2. F = ma, S = D/T3. I have looked into the equations of motion and Newton's Second Law, but cannot find a way to link these quantities.
 
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Nicaragua said:
1. A wooden block of mass m kg is at rest on a table, 1.6 metres from the edge. The block is pulled directly towards the edge by a horizontal string. The tension in the string has magnitude 0.2 mN. Calculate the time taken for the block to reach the edge of the table.2. F = ma, S = D/T3. I have looked into the equations of motion and Newton's Second Law, but cannot find a way to link these quantities.
Hi Nicaragua, Welcome to Physics Forums.

Please don't erase the template headings when you create your question statement.

Can you write out the basic kinematic equations that might be relevant? Look for ones relating acceleration, distance, and time. Using Newton's second law which you have stated, what is the acceleration of the block?
 
gneill said:
Hi Nicaragua, Welcome to Physics Forums.

Please don't erase the template headings when you create your question statement.

Can you write out the basic kinematic equations that might be relevant? Look for ones relating acceleration, distance, and time. Using Newton's second law which you have stated, what is the acceleration of the block?
Ok sorry I will keep the headers in next time. I don't know which equation to use when only given s = 1.6m and v = 0 m/s.
For F = ma, that means 0.2 mN / m = F, but we don't have mass either? I don't understand what equation to use when so little information is given.

Anyway, the equation linking acceleration distance and time is s = ut + 1/2 at^2
 
Nicaragua said:
Ok sorry I will keep the headers in next time. I don't know which equation to use when only given s = 1.6m and v = 0 m/s.
For F = ma, that means 0.2 mN / m = F, but we don't have mass either? I don't understand what equation to use when so little information is given.

Anyway, the equation linking acceleration distance and time is s = ut + 1/2 at^2

Okay, when the problem states that the force (tension) has a magnitude given by 0.2m N, it means that they're playing a little loosely with units. If the force is 0.2m Newtons then it will accelerate a mass m at the rate: a = 0.2m/m, or 0.2. Again, since the units are being used casually we can take that to be a = 0.2 m/s2 (where m here means meters, not the mass of the block)..

So you know the starting speed, the acceleration, and the distance to cover. It would appear that the extra Relevant Equation that you found will do the job nicely since its only other variable is time, which is what you want to find.
 
gneill said:
Okay, when the problem states that the force (tension) has a magnitude given by 0.2m N, it means that they're playing a little loosely with units.

Not as loosely as one might think. The OP states that the mass is m kg, meaning that m is a dimensionless number. The confusing part is when then quoting Newton's second law as F = ma...
 
Orodruin said:
Not as loosely as one might think. The OP states that the mass is m kg, meaning that m is a dimensionless number. The confusing part is when then quoting Newton's second law as F = ma...
So what equation would I use? I am missing too many quantities!?
 

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