#$&* course Phy 201 7 10/10 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
.............................................
Given Solution: `aA force Fnet acting on mass m results in acceleration a, where a = Fnet / m. We are given Fnet and a, so we can solve the equation to find m. Multiplying both sides by m we get a * m = Fnet / m * m so a * m = Fnet. Dividing both sides of this equation by a we have m = Fnet / a = 265 N / (2.30 m/s^2) = 115 (kg m/s^2) / (m/s^2) = 115 kg. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `qprin phy and gen phy problem 4.07 / 4.10a force to accelerate 7 g pellet to 125 m/s in .7 m barrel YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: vAve=(0+125m/s)/2=62.5m/s 62.5m/s=.7m/'dt 'dt=.11s a=125m/s/.11=11000m/s^2 F=m*a=.007g*11000=77N confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a** The initial velocity of the bullet is zero and the final velocity is 125 m/s. If we assume uniform acceleration (not necessarily the case but not a bad first approximation) the average velocity is (0 + 125 m/s) / 2 = 62.5 m/s and the time required for the trip down the barrel is .7 m / (62.5 m/s) = .011 sec, approx.. Acceleration is therefore rate of velocity change = `dv / `dt = (125 m/s - 0 m/s) / (.011 sec) = 11000 m/s^2, approx.. The force on the bullet is therefore F = m a = .007 kg * 11000 m/s^2 = 77 N approx. ** STUDENT COMMENT: I did my answer a different way and came up with a number just off of this. I calculated 78 and this solution shows an answer of 77, but I am positive that I did my work right. INSTRUCTOR RESPONSE: The results of my numerical calculations are always to be regarded as 'fuzzy'. The calculations are done mentally and there is often no intent to be exact. This at the very least encourages students to do the arithmetic and think about significant figures for themselves. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: prin: 10. A powerful motorcycle can produce an acceleration of 3.50 m/s^2 while traveling at 90.0 km/h. At that speed the forces resisting motion, including friction and air resistance, total 400 N. (Air resistance is analogous to air friction. It always opposes the motion of an object.) What force does the motorcycle exert backward on the ground to produce its acceleration if the mass of the motorcycle with rider is 245 kg? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution:
.............................................
Given Solution: The net force on the motorcycle must be F_net = m a = 245 kg * 3.5 m/s^2 = 870 kg m/s^2 = 870 Newtons, approximately. Intuitive solution: The road exerts a forward force which, combined with the opposing 400 Newtons of the air resistance, yields a net force of 870 Newtons. So the road must exert the 870 Newtons force, plus the additional 400 Newtons required to overcome air resistance, resulting in a net force of 1270 Newtons. This intuitive solution is good and useful. It captures the 'feel' of the situation. However it doesn't scale up well to more complex problems, which is why we need to construct a additional, more formal solution. More formal solution: If the direction of motion is regarded as positive and forward, the forces acting on the motorcycle include the forward-acting force between tire and road surface (specifically the frictional force of the road surface on the tire, which is equal and opposite to the frictional force exerted by the tire on the road surface), the backward-acting air resistance, the gravitational force pulling the motorcycle downward and the normal force of the road pushing it upward. The gravitational and normal forces are in this case equal and opposite, since the road is presumed to be level. The net force is therefore the sum of the forward, or positive, force of friction and the backward, or negative force of air resistance. We can therefore represent the net force as F_net = f_friction_on + f_air_resistance = f_friction_on + (-400 N), where f_friction_on is the frictional force between the tire and the road, acting on the motorcycle (as opposed to the frictional force produced by the motorcycle, which acts on the road). This net force is, as already noted, equal to about 870 Newtons. Setting our two expressions for net force equal, we find that f_friction_on - 400 N = 870 N so that f_friction_on = 870 N + 400 N = 1270 N. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: 13. The weight of an astronaut plus his space suit on the Moon is only 250 N. How much do they weigh on Earth? What is the mass on the Moon? On Earth? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: 250N=m*1.6m/s^2 160kg=m
.............................................
Given Solution: The weight of the astronaut on the moon is the gravitational force exerted on her by the Moon. That force is equal to the product of the acceleration of gravity on the Moon, and the mass of the astronaut: F_grav = mass * acceleration of gravity. The acceleration of gravity on the Moon is about 1.6 m/s^2. F_grav is the 250 N weight of the astronaut. So 250 N = mass * 1.6 m/s^2. We easily solve this equation to get mass = 250 N / (1.6 m/s^2) = 160 kg, approx. On Earth the acceleration of gravity is 9.8 m/s^2, so the weight of a 160 kg mass would be weight = 160 kg * 9.8 m/s^2 = 1570 N, approx. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `qgen phy 4.08 / 4.10b. A fish is being pulled upward. The breaking strength of the line holding the fish is 22 N. An acceleration of 2.5 m/s^2 breaks the line. What can we say about the mass of the fish? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Fnet=M*a T-M*g=M*2.5m/s^2 T=M*2.5m/s^2+M*g T=M*2.5m/s^2+M*9.8m/s^2 T=M*12.3m/s^2 M>22N/12.3m/s^2 M=1.8kg confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `aThe fish is being pulled upward by the tension, downward by gravity. The net force on the fish is therefore equal to the tension in the line, minus the force exerted by gravity. In symbols, Fnet = T - M g, where M is the mass of the fish. (We use capital M for the mass of the fish to distinguish the symbol for mass from the symbol m for meter). To accelerate a fish of mass M upward at 2.5 m/s^2 the net force must be Fnet = M a = M * 2.5 m/s^2. Combined with the preceding we have the condition M * 2.5 m/s^2 = T - M g so that to provide this force we require T = M * 2.5 m/s^2 + M g = M * 2.5 m/s^2 + M * 9.8 m/s^2 = M * 12.3 m/s^2. We know that the line breaks, so the tension must exceed the 22 N breaking strength of the line. So T > 22 N. Thus M * 12.3 m/s^2 > 22 N. Solving this inequality for m we get M > 22 N / (12.3 m/s^2) = 22 kg m/s^2 / (12.3 m/s^2) = 1.8 kg. The fish has a mass exceeding 1.8 kg. STUDENT QUESTION I had trouble understanding this question to begin with. I am a little confused on why the net force equals an acceleration of 12.3. INSTRUCTOR RESPONSE F_net = M a = M * 2.5 m/s^2, as expressed in the equation F_net = T - m g so that M * 2.5 m/s^2 = T - M g. It is the tension, not the net force, that ends up with a factor of 12.3 m/s^2: T = F_net + M g = M * 2.5 m/s^2 + M * 9.8 m/s^2, which is where the 12.3 m/s^2 comes from. Nothing actually accelerates at 12.3 m/s^2, just as nothing in this system accelerates at 9.8 m/s^2. 9.8 m/s^2 is the acceleration of gravity so M * 9.8 m/s^2 is the force exerted by gravity on the fish. M * 2.5 m/s^2 is the net force on the fish. To not only pull the fish upward against gravity, but to also accelerate it at 2.5 m/s^2, requires a tension force of M * 2.5 m/s^2 in addition to the force required to overcome gravity. Thus the tension force is M * 2.5 m/s^2 + M * 9.8 m/s^2 = M * 12.3 m/s^2. STUDENT QUESTION So the T does not really factor out of the equation it is just known that it is greater thatn or less than the Fnet? INSTRUCTOR RESPONSE Fnet is M * 2.5 m/s^2. We know that T = M * 12.3 m/s^2. We know that since the string breaks T is at least 22 N. So M * 12.3 m/s^2 is at least 22 N, and M must be at least 22 N / 12.3 m/s^2 = 1.8 kg. BRIEF SUMMARY The given solution tells you two things: The net force on the fish is T - M g, which is the sum of the tension and gravitational forces. The net force on the fish is M * 2.5 m/s^2, because of Newton's Second Law. We set these two expressions for the net force equal and solve for M. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `quniv phy 4.42 (11th edition 4.38) parachutist 55 kg with parachute, upward 620 N force. What are the weight and acceleration of parachutist? &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qDescribe the free body diagram you drew. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `aThe weight of the parachutist is 55 kg * 9.8 m/s^2 = 540 N, approx.. So the parachutist experiences a downward force of 540 N and an upward force of 620 N. Choosing upward as the positive direction the forces are -540 N and + 620 N, so the net force is -540 + 620 N = 80 N. Your free body diagram should clearly show these two forces, one acting upward and the other downward. The acceleration of the parachutist is a = Fnet / m = +80 N / (55 kg) = 1.4 m/s^2, approx.. STUDENT COMMENT I am having a hard time still yet understanding conversions, Ex) kg*m/s^2 = N, these hard for me to compute. they are not as hard since zi’ve been working with them but I am still having some trouble. INSTRUCTOR RESPONSE force = mass * acceleration, so the unit of force is the unit of mass * the unit of acceleration, i.e., kg * (m/s^2). We call this a Newton, but if you go back to the basic law you see why the basic unit is kg * m/s^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `quniv phy (4.34 10th edition) A fish hangs from a spring balance, which is in turn hung from the roof of an elevator. The balance reads 50 N when the elevator is accelerating at 2.45 m/s^2 in the upward direction. What is the net force on the fish when the balance reads 50 N? What is the true weight of the fish, under what circumstances will the balance read 30 N, and what will the balance read after the cable holding the fish breaks? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a** Weight is force exerted by gravity. Net force is Fnet = m * a. The forces acting on the fish are the 50 N upward force exerted by the cable and the downward force m g exerted by gravity. So m a = 50 N - m g, which we solve for m to get m = 50 N / (a + g) = 50 N / (2.45 m/s^2 + 9.8 m/s^2) = 50 N / 12.25 m/s^2 = 4 kg. If the balance reads 30 N then F_net = m a = 30 N - m g = 30 N - 4 kg * 9.8 m/s^2 = -9.2 N so a = -9.2 N / (4 kg) = -2.3 m/s^2; i.e., the elevator is accelerating downward at 2.3 m/s^2. If the cable breaks then the fish and everything else in the elevator will accelerate downward at 9.8 m/s^2. Net force will be -m g; net force is also Fbalance - m g. So -m g = Fbalance - m g and we conclude that the balance exerts no force. So it reads 0. ** STUDENT COMMENT I totally messed this problem up. I still have a hard time knowing how to setup my problems, but I understand solution INSTRUCTOR RESPONSE There are usually numerous ways to set up a given problem. In the case of this problem you want to start with Newton's Second Law, which you did. Having calculated the net force you could have set it equal to 50 N - m g, which would have given you 12.5 N = 50 N - m g with solution m = (50 N - 12.5 N) / g = (50 N - 12.5 N) / (9.8 m/s^2) = 4 kg, very approximately The symbolic equation would be m a = T - m g with solution m = T / (a + g) &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: `qSTUDENT QUESTION: I had trouble with the problems involving tension in lines. For example the Fish prob. Prob#9 A person yanks a fish out of water at 4.5 m/s^2 acceleration. His line is rated at 22 Newtons Max, His line breaks, What is the mass of the fish. Here's what I did. Sum of F = Fup + F down -22 N = 4.5 m/s^2 * m(fish) - 9.8 m/s^2 * m(fish) -22N = -5.3 m/s^2 m(fish) m(fish) = 4.2 kg I know its wrong, I just don't know what to do.I had the same problem with the elevator tension on problem 17. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a** Think in terms of net force. The net force on the fish must be Fnet = m a = m * 4.5 m/s^2. Net force is tension + weight = T - m g, assuming the upward direction is positive. So T - m g = m a and T = m a + m g. Factoring out m we have T = m ( a + g ) so that m = T / (a + g) = 22 N / (4.5 m/s^2 + 9.8 m/s^2) = 22 N / (14.3 m/s^2) = 1.8 kg, approx.. The same principles apply with the elevator. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!