course Phy 121 I saw on the Assignments page that Test 1 should be completed after this assignment. I thought we weren't supposed to take it until the end of this week's assignments. When should I take it? l??R????????=???Student Name:
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16:40:11 `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 --> mass on table= 5kg hanging mass = 2kg g = 9.8m/s^2 a = Fnet/m The force on the mass on the table is 5kg*9.8m/s^2 = 49 Newtons. Since the table is horizontal the mass has no acceleration. It is stable. Therefore, the force on this mass has no effect. The force on the hanging mass is 2kg * 9.8m/s^2 = 19.6 Newtons. Since this is the only force that moves the system, this is the only force on the system. Now we can find the acceleration. The net force is 19.6N and the total mass is 7kg. a = F/m = (19.6N)/(7kg) = 2.8 m/s^2
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16:41:06 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 --> I understand this problem.
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16:43:11 `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 --> The friction is between the mass on the table and the tabletop. Therefore, we use the force on this mass to calculate the force of friction. f = 49N * 0.10 = 4.9N Now we subtract the force lost to friction from the total force of the sytem that we found in the previous problem. Fnet = 19.6N - 4.9N = 14.7 N Then we can calculate the new acceleration which takes frictional force into consideration. a = F/m = 14.7 N / 7kg = 2.1 m/s^2
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16:43:34 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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RESPONSE --> Ok
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16:46:33 `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 --> The inline is descending therefore the slope is negative and below the positive x-axis. The incline makes an angle of 12 degrees with the negative y-axis. Therefore, the angle from the positive x-axis would be 90-12 = 78 degrees below the x-axis. We know that the vector is the magnitude of the force on the mass that sits on the incline. Therefore, the vector has a magnitude of 49 Newtons. Now we can find the force components with the vector length and the angle. x component = 49N * cos(-78) = 10.2 N y component = 49N * sin(-78) = -47.9 N From the first problem, we saw that the force of the y component must be zero. Therefore, the incline must exert 47.9 Newtons to counteract the -47.9 Newtons of the y component. The friction is 10% of the force on the mass at the top of the incline. The force on that mass is 47.9 Newtons, therefore the frictional force must be 4.79 Newtons. Now we can calculate the net force on the system by subtracting 4.79N of friction from 10.2N of x component force and the force from the hanging mass. 19.6N + 10.2N - 4.79N = 25 Newtons. By dividing the net force by the total mass, we get the acceleration of the system down the incline. a = Fnet/m = 25N/7kg = 3.57m/s^2
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16:47:55 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.
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?????~e?k??? assignment #020 020. `query 20 Physics I 07-03-2006
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18:31:09 Explain how we get the components of the resultant of two vectors from the components of the original vectors.
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RESPONSE --> If we have both components of two vectors, we simply add the x components to get the resultant x component and add the y components to get the resultant y component.
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18:32:20 ** 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 understand this problem.
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18:32:28 Explain how we get the components of a vector from its angle and magnitude.
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RESPONSE --> The x components of a vector can be found by multiplying the magnitude of the vector by the cosine of the angle the vector makes with the positive x-axis. The y component of the vector is found by multiplying the magnitude by the sine of that angle.
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18:32:42 ** 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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RESPONSE --> I understand this problem.
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18:35:36 prin phy, gen phy 7.02. Const frict force of 25 N on a 65 kg skiier for 20 sec; what is change in vel?
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RESPONSE --> f = 25N mass = 65kg `dt = 20 s a = F/m = `dv/`dt This can be rearranged to form the Impulse-Momentum Theorem: F*`dt = m*`dv (-25N)(20s) = (65kg)(`dv) -500N*s = 65kg*`dv `dv = -7.69 m/s
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18:35:59 If the direction of the velocity is taken to be positive, then the directio of the frictional force is negative. A constant frictional force of -25 N for 20 sec delivers an impulse of -25 N * 20 sec = -500 N sec in the direction opposite the velocity. By the impulse-momentum theorem we have impulse = change in momentum, so the change in momentum must be -500 N sec. The change in momentum is m * `dv, so we have m `dv = impulse and `dv = impulse / m = -500 N s / (65 kg) = -7.7 N s / kg = -7.7 kg m/s^2 * s / kg = -7.7 m/s.
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RESPONSE --> I understand this problem.
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18:36:12 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 --> This problem does not apply to me.
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18:36:16 **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 -->
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18:36:26 **** 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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RESPONSE --> This problem does not apply to me.
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18:36:30 ** 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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RESPONSE -->
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