course Phy 201 I feel like I am doing horribly on these assignments, but I am learning from them. 螲y蚨岡厨岃甞渎燲否詬assignment #010
......!!!!!!!!...................................
14:13:19 `questionNumber 100000 `q001. If a block of mass 10 kg accelerates at 2 m/s^2, then what net force is acting on the block?
......!!!!!!!!...................................
RESPONSE --> F=m*a =10*2 =20 Newtons confidence assessment: 3
.................................................
......!!!!!!!!...................................
14:13:27 `questionNumber 100000 The net force on the block is the product F = m * a of its 10 kg mass and its 2 m/s^2 acceleration. The net force is therefore F = 10 kg * 2 m/s^2 = 20 kg * m / s^2. The unit of force, which is the product of a quantity in kg and another quantity in m/s^2, is just the algebraic product kg * m/s^2 of these two units. This unit, the kg * m / s^2, is called a Newton. So the net force is 20 Newtons.
......!!!!!!!!...................................
RESPONSE --> I understand. self critique assessment: 3
.................................................
......!!!!!!!!...................................
14:16:00 `questionNumber 100000 `q002. How much force must be exerted by someone pulling on it to accelerate a 10 kg object at 2 m/s^2?
......!!!!!!!!...................................
RESPONSE --> 10*2 20 Newtons confidence assessment: 2
.................................................
......!!!!!!!!...................................
14:16:12 `questionNumber 100000 This depends on what forces might be resisting the acceleration of the object. If the object is accelerating on a surface of some type, then there is a good chance that a frictional force is opposing the motion. If the object is being pulled upward against the force of gravity, then more force is required then if it is sliding along a low-friction horizontal surface. If it is being pulled downhill, the force exerted by gravity has a component in the direction of motion and perhaps even less force is required. However, in every case the net force, which is the sum of all the forces acting on the object, must be 20 Newtons. The person pulling on the object must exert exactly enough force that the net force will be 20 Newtons.
......!!!!!!!!...................................
RESPONSE --> I understand. self critique assessment: 3
.................................................
......!!!!!!!!...................................
14:18:08 `questionNumber 100000 `q003. If friction exerts a force of 10 Newtons in the direction opposite the motion of a 10 kg object, then how much force must be exerted by someone pulling on it to accelerate the 10 kg object at 2 m/s^2, with the acceleration in the same direction as the motion?
......!!!!!!!!...................................
RESPONSE --> F=10*2 =20 20+10 (opposing Newtons)= 30 total Newtons confidence assessment: 2
.................................................
......!!!!!!!!...................................
14:18:26 `questionNumber 100000 Since the 10 Newton frictional force is in the direction opposite to motion, and since the acceleration is in the same direction as the motion, the frictional force is opposed to the accelerating force. If the direction of motion is taken as positive, then the frictional force will be in the negative direction and can be denoted fFrict = - 10 Newtons. To achieve the given acceleration the net force on the object must be net force = 10 kg * (+2 m/s^2) = +20 kg * m/s^2 = +20 Newtons. In order to achieve the +20 Newton net force when there is already a frictional force of -10 Newtons, it should be clear that a force of +30 Newtons is required. This can be thought of as 10 Newtons to overcome friction and another 20 Newtons to achieve the required net force.
......!!!!!!!!...................................
RESPONSE --> I understand. self critique assessment: 3
.................................................
......!!!!!!!!...................................
14:27:58 `questionNumber 100000 `q004. How can we write an equation to solve this problem? Hint: What equation would relate the net force Fnet, the force F exerted by the person and the force fFrict of friction?
......!!!!!!!!...................................
RESPONSE --> I honestly don't know and the hint, sadly, doesn't even help me! confidence assessment: 0
.................................................
......!!!!!!!!...................................
14:30:21 `questionNumber 100000 If Fnet is the net force and F the force actually exerted by the person, then Fnet = F + fFrict. That is, the net force is the sum of the force exerted by the person and the frictional force. We know that Fnet is +20 Newtons and fFrict is -10 Newtons, so we have the equation 20 Newtons = F + (-10 Newtons). Solving for F we see that F = 20 Newtons + 10 Newtons = 30 Newtons.
......!!!!!!!!...................................
RESPONSE --> This makes sense since the definition of net force is the sum of all forces, even if the forces aren't necessarily going the same way the whole time (that is, their forces aren't working in the same direction). self critique assessment: 2
.................................................
......!!!!!!!!...................................
15:22:16 `questionNumber 100000 `q005. If a constant net force of 12 Newtons acts on a cart of mass 6 kg, then at what rate does the velocity of the cart change?
......!!!!!!!!...................................
RESPONSE --> I don't know. confidence assessment: 0
.................................................
......!!!!!!!!...................................
15:23:39 `questionNumber 100000 The velocity of the cart will change at a rate a which is related to the net force and the mass by Fnet = m * a. Thus a = Fnet / m = 12 Newtons / (6 kg) = 12 kg * m/s^2 / (6 kg) = 2 m/s^2. We note that the force unit Newtons is broken down to its fundamental units of kg * m / s^2 in order to perform the unit calculation. Dividing kg * m / s^2 by kg we have (kg / kg) * m/s^2 = m/s^2. It is important to always do the unit calculations. This habit avoids a large number of errors and also can be used to reinforce our understanding of the relationships in a problem or situation.
......!!!!!!!!...................................
RESPONSE --> Wow that is a convenient formula then. This isn't something that I didn't understand--I just didn't know it. So now I do, and I hope to be able to apply it the next time the question is put forth. self critique assessment: 2
.................................................
......!!!!!!!!...................................
15:26:02 `questionNumber 100000 `q006. If a force of 50 Newtons is exerted in the direction of the object's motion by a person, on a 20 kg object, and if friction exerts a force of 10 Newtons opposed to the direction of motion, then what will be the acceleration of the object?
......!!!!!!!!...................................
RESPONSE --> The force necessary will be 50+10 since 10 is going in the opposite direction, yielding F=60 F=m*a 60=20*a 30 kg/s^2=a confidence assessment: 2
.................................................
......!!!!!!!!...................................
15:27:52 `questionNumber 100000 The object will accelerate at a rate determined by Newton's Second Law, Fnet = m * a. The acceleration will therefore be a = Fnet / m. The net force on the object will be the sum of the 50 Newton force in the direction of motion and the 10 Newton force opposed to the direction of motion. If we take the direction of motion as positive, then the net force is Fnet = 50 N - 10 N = 40 N. It follows that the acceleration is a = Fnet / m = 40 N / (20 kg) = 40 kg m/s^2 / (20 kg) = 2 m/s^2.
......!!!!!!!!...................................
RESPONSE --> I see that instead of adding 10 I should have subtracted it. This time, the question was not asking ""how much force is necessary to do x,""; instead, it was saying, ""how much total force is there"" and that number was what should have been entered into the equation. Besides entering the wrong number for Fnet, I understand the rest of the procedure. self critique assessment: 2
.................................................
......!!!!!!!!...................................
15:29:08 `questionNumber 100000 `q007. If a force of 50 Newtons is exerted opposite to the direction of the object's motion by a person, on a 20 kg object, and if friction exerts a force of 10 Newtons opposed to the direction of motion, then what will be the acceleration of the object?
......!!!!!!!!...................................
RESPONSE --> (50-10) = 40 Fnet=m*a 40=20kg*a 2 kg/s^2=a confidence assessment: 3
.................................................
......!!!!!!!!...................................
15:31:26 `questionNumber 100000 If we take the direction of motion to be positive, then since both the 50 Newton force and the 10 Newton force are opposed to the direction of motion the net force must be net force = -50 Newtons - 10 Newtons = -60 Newtons. The acceleration of the object will therefore be a = Fnet / m = -60 Newtons / (10 kg) = -60 kg * m/s^2 / (20 kg) = -3 m/s^2. The fact that the acceleration is opposed to the direction of motion indicates that the object will be slowing down. The force exerted by the person, being in the direction opposite to that of the motion, is seen to be a retarding force, as is friction. So in this case the person is aided by friction in her apparent goal of stopping or at least slowing the object.
......!!!!!!!!...................................
RESPONSE --> Well, this time I at least knew to subtract; unfortunately, I didn't clue in to the fact that the object was moving in a negative direction. I feel that I understand the equation still, but I need a lot more practice with this ""moving in the direction of, against, etc."" self critique assessment: 2
.................................................
......!!!!!!!!...................................
15:38:19 `questionNumber 100000 `q008. If a 40 kg object is moving at 20 m/s, then how long will a take a net force of 20 Newtons directed opposite to the motion of the object to bring the object to rest?
......!!!!!!!!...................................
RESPONSE --> We can substitute in the 20 m/s for the Fnet, leaving 20 m/s=40 kg*a 20/40=.5 Beyond this point, I do not know how to solve the problem. confidence assessment: 0
.................................................
......!!!!!!!!...................................
15:41:36 `questionNumber 100000 The force on the object is in the direction opposite its motion, so if the direction of motion is taken to be positive the force is in the negative direction. We therefore write the net force as Fnet = -20 Newtons. The acceleration of the object is therefore a = Fnet / m = -20 Newtons / 40 kg = -20 kg * m/s^2 / (40 kg) = -.5 m/s^2. We can therefore describe uniformly accelerated motion of the object as v0 = 20 m/s, vf = 0 (the object comes to rest, which means its velocity ends up at 0), a = -.5 m/s^2. We can then reason out the time required from the -20 m/s change in velocity and the -.5 m/s^2 acceleration, obtaining `dt = 40 seconds. We can confirm this using the equation vf = v0 + a `dt: Solving for `dt we obtain `dt = (vf - v0) / a = (0 m/s - 20 m/s) / (-.5 m/s^2) = -20 m/s / (-.5 m/s^2) = 40 m/s * s^2 / m = 40 s.
......!!!!!!!!...................................
RESPONSE --> I understand. I just need to use the F=m*a formula, apply it to the question directly, adn then use the acceleration information derived from it to complete a uniform acceleration problem. self critique assessment: 2
.................................................
......!!!!!!!!...................................
15:44:48 `questionNumber 100000 `q009. If we wish to bring an object with mass 50 kg from velocity 10 m/s to velocity 40 m/s in 5 seconds, what constant net force would be required?
......!!!!!!!!...................................
RESPONSE --> a=vf-vo/'dt =10m/s-40m/s/5 seconds =-7.8 m/s^2 Fnet=m*a =50kg*-7.8 m/s^2 =-390 Newtons confidence assessment: 1
.................................................
......!!!!!!!!...................................
15:46:18 `questionNumber 100000 The net force would be Fnet = m * a. The acceleration of the object would be the rate which its velocity changes. From 10 m/s to 40 m/s the change in velocity is +30 m/s; to accomplish this in 5 seconds requires average acceleration 30 m/s / (5 s) = 6 m/s^2. Thus the net force required is Fnet = 50 kg * 6 m/s^2 = 300 kg m/s^2 = 300 Newtons.
......!!!!!!!!...................................
RESPONSE --> I calculated the acceleration wrong, but for the first time I understood before seeing the solution how I was supposed to go about solving it. self critique assessment: 2
.................................................
......!!!!!!!!...................................
15:55:20 `questionNumber 100000 `q010. If a constant net force of 50 Newtons brings an object to rest in four seconds from an initial velocity of 8 meters/second, then what must be the mass of the object?
......!!!!!!!!...................................
RESPONSE --> I know that we have to find acceleration before we can do anything else. Unfortunately, I am stuck on how to do that. confidence assessment: 2
.................................................
......!!!!!!!!...................................
16:00:50 `questionNumber 100000 We know the net force and we have the information required to calculate the acceleration. We will therefore be able to find the mass using Newton's Second Law Fnet = m * a. We first find the acceleration. The change in velocity from 8 m/s to rest is -8 m/s, and this occurs in 4 seconds. The acceleration is therefore -8 m/s / (4 s) = -2 m/s^2. The 50 Newton net force must be in the same direction as the acceleration, so we have Fnet = -50 Newtons. We obtain the mass by solving Newton's Second Law for m: m = Fnet / a = -50 N / (-2 m/s^2) = -50 kg m/s^2 / (-2 m/s^2) = 25 kg.
......!!!!!!!!...................................
RESPONSE --> How easy. This stuff makes me feel so stupid. We find the acceleration by taking the change in the initial velocity and what we hope is final velocity 0. The initial velocity we assume to be 0, so 0-8 = -8 m/s. This divided by the time according to the aAve formula gives us -2m/s^2. The rest of the formula is fairly simple to fill in. self critique assessment: 2
................................................."