course Phy 201 Here is my work for Assignment 20. I will be taking my test # 1 on Tuesday night at the college. I am courious as to how I did on my quiz. I also emailed you about my grades on the Blackboard page and how they don't tell you anything but the class average. I would like to know what my current standing is at the end of the semester and if it is good enough to get me a
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19:30:18 `q001. Note that this assignment contains 3 questions. . A 5 kg block rests on a tabletop. A string runs horizontally from the block over a pulley of negligible mass and with negligible friction at the edge of the table. There is a 2 kg block hanging from the string. If there is no friction between the block in the tabletop, what should be the acceleration of the system after its release?
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RESPONSE --> I know that the 2 kg block will fall at the acceleration of free fall which will be 9.8 m/s^2 and the 5 kg block will actually work as friction against this but I am not sure how to get the answer?
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19:33:18 Gravity exerts a force of 5 kg * 9.8 meters/second = 49 Newtons on the block, but presumably the tabletop is strong enough to support the block and so exerts exactly enough force, 49 Newtons upward, to support the block. The total of this supporting force and the gravitational force is zero. The gravitational force of 2 kg * 9.8 meters/second = 19.6 Newtons is not balanced by any force acting on the two mass system, so we have a system of total mass 7 kg subject to a net force of 19.6 Newtons. The acceleration of this system will therefore be 19.6 Newtons/(7 kg) = 2.8 meters/second ^ 2.
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RESPONSE --> Ok I see what you did and it was to add the total mass of the system and then divide the Newtons pulling on the blocks in a downward motion. This would then give us the acceleration of the whole system. I have been away and I need a little refresher.
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19:38:01 `q002. Answer the same question as that of the previous problem, except this time take into account friction between the block in the tabletop, which exerts a force opposed to motion which is .10 times as great as the force between the tabletop and the block. Assume that the system slides in the direction in which it is accelerated by gravity.
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RESPONSE --> Well it has been established that the force in the block on the tabletop is 49 Newtons. So if we have a force opposing the direction of motion then that would be 49 Newtons * .10 Newtons = 4.9 Newtons. Now if this force is opposing the direction of the motion then this would mean that we would need to subtract 4.9 Newtons from the 19.6 Newtons of the system. This would give us a total of 14.7 Newtons. And if we then divide 14.7 Newtons by 7 kg we get 2.1 m/s^2.
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19:38:09 Again the weight of the object is exactly balance by the upward force of the table on the block. This force has a magnitude of 49 Newtons. Thus friction exerts a force of .10 * 49 Newtons = 4.9 Newtons. This force will act in the direction opposite that of the motion of the system. It will therefore be opposed to the 19.6 Newton force exerted by gravity on the 2 kg object. The net force on the system is therefore 19.6 Newtons -4.9 Newtons = 14.7 Newtons. The system will therefore accelerate at rate a = 14.7 Newtons/(7 kg) = 2.1 meters/second ^ 2.
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19:51:33 `q003. Answer the same question as that of the preceding problem, but this time assume that the 5 kg object is not on a level tabletop but on an incline at an angle of 12 degrees, and with the incline descending in the direction of the pulley.
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RESPONSE --> Ok this may not be right but this is how I see it. Gravity is now exerted a force on the 5 kg block on the tabletop and it is equal to that of a 12 degree angle in freefall. So that would mean that the acceleration of the 5 kg block is now 9.8 m/s^2 * .1333 = 1.31 m/s^2. I got the .1333 by considering a 90 degree angle as 100 % of freefall and then dividing 12 by 90 to get that a 12 degree angle is 13 % of freefall.
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19:55:40 In this case you should have drawn the incline with the x axis pointing down the incline and the y axis perpendicular to the incline. Thus the x axis is directed 12 degrees below horizontal. As a result the weight vector, rather than being directed along the negative y axis, lies in the fourth quadrant of the coordinate system at an angle of 12 degrees with the negative y axis. So the weight vector makes an angle of 270 degrees + 12 degrees = 282 degrees with the positive x axis. The weight vector, which has magnitude 5 kg * 9.8 meters/second ^ 2 = 49 Newtons, therefore has x component 49 Newtons * cosine (282 degrees) = 10 Newtons approximately. Its y component is 49 Newtons * sine (282 degrees) = -48 Newtons, approximately. The incline exerts sufficient force that the net y component of the force on the block is zero. The incline therefore exerts a force of + 48 Newtons. Friction exerts a force which is .10 of this force, or .10 * 48 Newtons = 4.8 Newtons opposed to the direction of motion. Assuming that the direction of motion is down the incline, frictional force is therefore -4.8 Newtons in the x direction. The weight component in the x direction and the frictional force in this direction therefore total 10 Newtons + (-4.8 Newtons) = + 5.2 Newtons. This force tends to accelerate the system in the same direction as does the weight of the 2 kg mass. This results in a net force of 5.2 Newtons + 19.6 Newtons = 24.8 Newtons on the 7 kg system. The system therefore accelerates at rate {} a = (24.8 Newtons) / (7 kg) = 3.5 meters/second ^ 2, approximately.
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RESPONSE --> Ok I see how you got this, and as you can see my way got me close but still not correct. I will practice this one some more.
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]v`yv assignment #020 yE܃֭ Physics I 12-12-2005
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20:33:25 Explain how we get the components of the resultant of two vectors from the components of the original vectors.
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RESPONSE --> Well what we would need to do is angle and the vector length and do some math. First we would need to take the vector length and multiply it by the cos of the angle to get the x component. Then we would need to take the vector length and multiply it by the sin of the angle to get the y component.
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20:34:17 ** If we add the x components of the original two vectors we get the x component of the resultant. If we add the y components of the original two vectors we get the y component of the resultant. **
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RESPONSE --> I guess I didn't understand the problem to say that but you can see what I meant and that I understand this.
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20:34:43 Explain how we get the components of a vector from its angle and magnitude.
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RESPONSE --> Ok that is what I explained a question ago.
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20:34:45 ** To get the y component we multiply the magnitude by the sine of the angle of the vector (as measured counterclockwise from the positive x axis). To get the x component we multiply the magnitude by the cosine of the angle of the vector (as measured counterclockwise from the positive x axis). **
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20:37:22 gen phy #7.12 23 g bullet 230 m/s 2-kg block emerges at 170 m/s speed of block
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RESPONSE --> I am not sure how to set this up.
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20:40:40 **STUDENT SOLUTION: Momentum conservation gives us m1 v1 + m2 v2 = m1 v1' + m2 v2' so if we let v2' = v, we have (.023kg)(230m/s)+(2kg)(0m/s) = (.023kg)(170m/s)+(2kg)(v). Solving for v: (5.29kg m/s)+0 = (3.91 kg m/s)+(2kg)(v) .78kg m/s = 2kg * v v = 1.38 kg m/s / (2 kg) = .69 m/s. INSTRUCTOR COMMENT: It's probably easier to solve for the variable v2 ': Starting with m1 v1 + m2 v2 = m1 v1 ' + m2 v2 ' we add -m1 v1 ' to both sides to get m1 v1 + m2 v2 - m1 v1 ' = m2 v2 ', then multiply both sides by 1 / m2 to get v2 ` = (m1 v1 + m2 v2 - m1 v1 ' ) / m2. Substituting for m1, v1, m2, v2 we will get the result you obtained.**
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RESPONSE --> Ok I see how this is done. I guess it is the equation that I was not hip on. I will be sure to have this down by the test.
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20:40:42 **** Univ. 8.70 (8.68 10th edition). 8 g bullet into .992 kg block, compresses spring 15 cm. .75 N force compresses .25 cm. Vel of block just after impact, vel of bullet?
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20:40:44 ** The spring ideally obeys Hook's Law F = k x. It follows that k = .75 N / .25 cm = 3 N / cm. At compression 15 cm the potential energy of the system is PE = .5 k x^2 = .5 * 3 N/cm * (15 cm)^2 = 337.5 N cm, or 3.375 N m = 3.375 Joules, which we round to three significant figures to get 3.38 J. The KE of the 1 kg mass (block + bullet) just after impact is in the ideal case all converted to this PE so the velocity of the block was v such that .5 m v^2 = PE, or v = sqrt(2 PE / m) = sqrt(2 * 3.38 J / ( 1 kg) ) = 2.6 m/s, approx. The momentum of the 1 kg mass was therefore 2.6 m/s * .992 kg = 2.6 kg m/s, approx., just after collision with the bullet. Just before collision the momentum of the block was zero so by conservation of momentum the momentum of the bullet was 2.6 kg m/s. So we have mBullet * vBullet = 2.6 kg m/s and vBullet = 2.6 kg m/s / (.008 kg) = 330 m/s, approx. **
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