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One block (3.3 kg) is on top of a table and is connected by a massless cord over a frictionless massley pulley to another block(2 kg) which is hanging down over the right side of the table. The system starts at rest. When the masses have moved a distance of 0.431 m, their speed is 1.25 m/s. The acceleration of gravity is 9.8 m/s^2.

What is the coefficient of the friction between m2 (m subscript 2)and the table (answer in N)?

What is the magnitude of the tension in the cord (answer in N)?

Thanks in advance for your input.......

2006-09-30 04:06:34 · 2 answers · asked by Dee 4 in Science & Mathematics Physics

2 answers

There are a few key ideas to remember in this problem. First, the force that causes the acceleration is due to the falling block (2 kg x 9.8 m/s^2 = 19.6 N), and the total mass accelerating is (3.3 kg + 2 kg = 5.3 kg). Fiction opposes the acceleration.

The value for the actual acceleration in this case can be found using the formula Vf^2 = Vi^2 + 2ad. Rearrange the equation to solve for a, using the known variables (Vi=0 m/s, Vf = 1.25 m/s and d=0.431 m) You should get a=1.81 m/s^2.

In order to find the coefficient of friction (u=Ff/Fn) you need to find the force of friction. To do this, use Newton's 2nd law, F=ma. (remember that F is the net force, which is the force of gravity minus the force of friction.) F = 5.3kg * 1.81 m/s^2 = 9.6 N.

Since 9.6 N represents the NET force, find the force friction using the formula above (Fg - Ff = 9.6N). Since Fg is 2kg * -9.8m/s^2, Ff works out to be about 10N.

Now you should be able to solve for the coefficient of friction, u= Ff/Fn = 10 N/ (3.3kg *9.8m/s^2) = 0.31.

The tension in the cord is equal to the Net force, which is just 9.6 N, as shown above.

Hope this helps!!!

2006-09-30 04:41:38 · answer #1 · answered by cushdogjr 3 · 0 0

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2016-10-15 09:12:46 · answer #2 · answered by ? 4 · 0 0

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