Question

upward direction as positive. The total mass M of the system (car plus passengers) is equal to 1,660kg. Using the particle in equilibrium model, apply Newton's second law to the car: Solve for T : Use P=vec(F)vec(v) and that vec(T) is in the same direction as vec(v) to find the power: Substitute numerical values: sum F_(y)=T-f-Mg=0 T=Mg+f P=vec(T)vec(v)=TV=(Mg+f)v P=[(1,660(kg))(9.80(m)/(s^(2)))+(4,050(N))](2.80(m)/(s)) =,W w (B) What power must the motor deliver at the instant the speed of the elevator is v if the motor is designed to provide the elevator car with an upward acceleration of 2.00(m)/(s^(2)) ? Conceptualize In this case, the motor must supply the force of magnitude T that pulls the elevator car upward with an increasing speed. We expect that more power will be required to do that than in part (A) because the motor must now perform the additional task of accelerating the car. Categorize In this case, we model the elevator car as a particle under a net force because it is accelerating. Analyze Apply Newton's second law to the car: Solve for T : Use P=vec(F)*vec(v) to obtain the required power: Substitute numerical values: sum F_(y)=T-f-Mg=Ma T=M(a+g)+f P=TV=[M(a+g)+f]v P=[(1,660(kg))(2.00(m)/(s^(2))+9.80(m)/(s^(2)))+4,050(N)]v =, imes v Finalize To compare with part (A), let v=2.80(m)/(s), giving a power of P=T(2.80(m)/(s))= which is larger than the power found in part (A), as expected. upward direction as positive. The total mass M of the system (car plus passengers) is equal to 1,660 kg Using the particle in equilibrium model, apply Newton's second law to the car: =T -f -Mg =0 Solve for T: T = Mg + f Use P = F . V and that T is in the same direction as v to find the power: P=.=Tv=(Mg+fv Substitute numerical values: P = [(1,660 kg)(9.80 m/s2) + (4,050 N)](2.80 m/s) 56,862.4 W (B) what power must the motor deliver at the instant the speed of the elevator is v if the motor is designed to provide the elevator car with an upward acceleration of 2.00 m/s2? Conceptualize In this case, the motor must supply the force of magnitude T that pulls the elevator car upward with an increasing speed. We expect that more power will be required to do that than in part (A) because the motor must now perform the additional task of accelerating the car. Categorize In this case, we model the elevator car as a particle under a net force because it is accelerating. Analyze Apply Newton's second law to the car: Fy=T -f -Mg=Ma Solve for T: T = M(a + g) + f Use P = F . V to obtain the required power: P=Tv=[Ma+g)+f]v Substitute numerical values: P = [(1,660 kg)(2.00 m/s2 + 9.80 m/s2) + 4,050 N]v =56 X v Finalize To compare with part (A), let v = 2.80 m/s, giving a power of P=T2.80m/s)=56,862.4 Ix w which is larger than the power found in part (A), as expected.

Name: upward direction as positive the total mass m of the system car plus passengers is equal to 1660kg using the particle in equilibrium model apply newtons second law to the car solve for t use 80677
Uploaded: 2021-04-06T22:41:40-08:00
Duration: 47 min 52 s
Channel: Adam Schade
Description: upward direction as positive the total mass m of the system car plus passengers is equal to 1660kg using the particle in equilibrium model apply newtons second law to the car solve for t use 80677

          upward direction as positive. The total mass M of the system (car plus passengers) is equal to 1,660kg.
Using the particle in equilibrium model,
apply Newton's second law to the car:
Solve for T :
Use P=vec(F)*vec(v) and that vec(T) is in the same
direction as vec(v) to find the power:
Substitute numerical values:
sum F_(y)=T-f-Mg=0
T=Mg+f
P=vec(T)*vec(v)=TV=(Mg+f)v
P=[(1,660(kg))(9.80(m)/(s^(2)))+(4,050(N))](2.80(m)/(s))
=,W
w
(B) What power must the motor deliver at the instant the speed of the elevator is v if the motor is
designed to provide the elevator car with an upward acceleration of 2.00(m)/(s^(2)) ?
Conceptualize In this case, the motor must supply the force of magnitude T that pulls the elevator car
upward with an increasing speed. We expect that more power will be required to do that than in part (A)
because the motor must now perform the additional task of accelerating the car.
Categorize In this case, we model the elevator car as a particle under a net force because it is
accelerating.
Analyze Apply Newton's second law to the
car:
Solve for T :
Use P=vec(F)*vec(v) to obtain the required power:
Substitute numerical values:
sum F_(y)=T-f-Mg=Ma
T=M(a+g)+f
P=TV=[M(a+g)+f]v
P=[(1,660(kg))(2.00(m)/(s^(2))+9.80(m)/(s^(2)))+4,050(N)]v
=,	imes v
Finalize To compare with part (A), let v=2.80(m)/(s), giving a power of
P=T(2.80(m)/(s))=
which is larger than the power found in part (A), as expected.
upward direction as positive. The total mass M of the system (car plus passengers) is equal to 1,660 kg
Using the particle in equilibrium model, apply Newton's second law to the car:
=T -f -Mg =0
Solve for T:
T = Mg + f
Use P = F . V and that T is in the same direction as v to find the power:
P=.=Tv=(Mg+fv
Substitute numerical values:
P = [(1,660 kg)(9.80 m/s2) + (4,050 N)](2.80 m/s) 56,862.4 W
(B) what power must the motor deliver at the instant the speed of the elevator is v if the motor is
designed to provide the elevator car with an upward acceleration of 2.00 m/s2?
Conceptualize In this case, the motor must supply the force of magnitude T that pulls the elevator car upward with an increasing speed. We expect that more power will be required to do that than in part (A) because the motor must now perform the additional task of accelerating the car.
Categorize In this case, we model the elevator car as a particle under a net force because it is accelerating.
Analyze Apply Newton's second law to the car:
Fy=T -f -Mg=Ma
Solve for T:
T = M(a + g) + f
Use P = F . V to obtain the required power:
P=Tv=[Ma+g)+f]v
Substitute numerical values:
P = [(1,660 kg)(2.00 m/s2 + 9.80 m/s2) + 4,050 N]v =56 X v
Finalize To compare with part (A), let v = 2.80 m/s, giving a power of P=T2.80m/s)=56,862.4 Ix w
which is larger than the power found in part (A), as expected.

upward direction as positive the total mass m of the system car plus passengers is equal to 1660kg using the particle in equilibrium model apply newtons second law to the car solve for t use 80677

Added by Adriana C.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Adam Schade

04/06/2021

Step 1

Step 1: Apply Newton's second law to the car: The sum of the forces in the y-direction is given by: \sum F_(y) = T - f - Mg = Ma Solve for T: T = M(a + g) + f Show more…

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upward direction as positive. The total mass M of the system (car plus passengers) is equal to 1,660kg. Using the particle in equilibrium model, apply Newton's second law to the car: Solve for T : Use P=vec(F)*vec(v) and that vec(T) is in the same direction as vec(v) to find the power: Substitute numerical values: sum F_(y)=T-f-Mg=0 T=Mg+f P=vec(T)*vec(v)=TV=(Mg+f)v P=[(1,660(kg))(9.80(m)/(s^(2)))+(4,050(N))](2.80(m)/(s)) =,W w (B) What power must the motor deliver at the instant the speed of the elevator is v if the motor is designed to provide the elevator car with an upward acceleration of 2.00(m)/(s^(2)) ? Conceptualize In this case, the motor must supply the force of magnitude T that pulls the elevator car upward with an increasing speed. We expect that more power will be required to do that than in part (A) because the motor must now perform the additional task of accelerating the car. Categorize In this case, we model the elevator car as a particle under a net force because it is accelerating. Analyze Apply Newton's second law to the car: Solve for T : Use P=vec(F)*vec(v) to obtain the required power: Substitute numerical values: sum F_(y)=T-f-Mg=Ma T=M(a+g)+f P=TV=[M(a+g)+f]v P=[(1,660(kg))(2.00(m)/(s^(2))+9.80(m)/(s^(2)))+4,050(N)]v =, imes v Finalize To compare with part (A), let v=2.80(m)/(s), giving a power of P=T(2.80(m)/(s))= which is larger than the power found in part (A), as expected. upward direction as positive. The total mass M of the system (car plus passengers) is equal to 1,660 kg Using the particle in equilibrium model, apply Newton's second law to the car: =T -f -Mg =0 Solve for T: T = Mg + f Use P = F . V and that T is in the same direction as v to find the power: P=.=Tv=(Mg+fv Substitute numerical values: P = [(1,660 kg)(9.80 m/s2) + (4,050 N)](2.80 m/s) 56,862.4 W (B) what power must the motor deliver at the instant the speed of the elevator is v if the motor is designed to provide the elevator car with an upward acceleration of 2.00 m/s2? Conceptualize In this case, the motor must supply the force of magnitude T that pulls the elevator car upward with an increasing speed. We expect that more power will be required to do that than in part (A) because the motor must now perform the additional task of accelerating the car. Categorize In this case, we model the elevator car as a particle under a net force because it is accelerating. Analyze Apply Newton's second law to the car: Fy=T -f -Mg=Ma Solve for T: T = M(a + g) + f Use P = F . V to obtain the required power: P=Tv=[Ma+g)+f]v Substitute numerical values: P = [(1,660 kg)(2.00 m/s2 + 9.80 m/s2) + 4,050 N]v =56 X v Finalize To compare with part (A), let v = 2.80 m/s, giving a power of P=T2.80m/s)=56,862.4 Ix w which is larger than the power found in part (A), as expected.

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Transcript

00:01 Okay, so here's the problem we're going to be working on.

00:04 A 50 kilogram man standing on a spring scale, taking an elevator right down.

00:11 For the first part of the problem, it says that the elevator starts from rest, and it takes four seconds to reach a speed of 8 meters per second.

00:24 So here's a little picture just to show what's going on.

00:31 The guy standing on a spring scale in an elevator and the elevator is going down.

00:39 Good thing to do whenever you're doing a physics problem is to write up above all of the values and constants that it gives you.

00:51 So it gives us the mass of the person is 50 kilograms.

00:55 The initial velocity is zero because it starts from rest.

01:00 The final velocity of this.

01:02 First part is 8 meters per seconds and the time it takes for this first part is 4 seconds and i'm going to be working on each of the parts separately so for this first question it asks us what the acceleration was or at least the average acceleration for this first part so in order to solve this let's look at our kinematic equations that you should hopefully have written down somewhere and it looks like the equation that we're going to want to use is going to be the final velocity, what we're calling v1 is equal to the initial velocity plus the acceleration times.

02:02 And since we're solving for acceleration, let's move everything except for the acceleration to one side.

02:11 So we're going to be, let's see, let's subtract the initial velocity, so that will end up being v1 minus v initial is equal to a1 times t1.

02:33 Let's divide by t1, and we should get a1 is equal to v1 minus v initial, all over, t1 and looks like each of these variables are given up here so let's just plug them in so let's see the final velocity v1 is going to be eight meters per seconds subtracted by the initial velocity which is zero meters per seconds divided by the time which is four seconds let's simplify this that'd be eight meters per seconds over four seconds which ends up being two meters per second squared always helps to double check your units as well it looks like we're have meters per seconds and we're dividing it by seconds so that would be meters over seconds time seconds or meters per second squared so our units check out there cool let's look at the second part of the problem it says how far does the elevator travel during this time so it gets another good thing to remember is that uh the x initial or the initial starting point for this problem is going to be zero not always but if it doesn't specify distance then it's best to always assume that the initial distance is zero that's asking us for the distance of traveler travel travels during this first part so let's look back at our kinematic equations and it looks like the equation we're going to want to use is going to be x final or displacement final is equal to x initial plus v initial times t1 plus one half a1 times t1 squared okay and there's a reason that i'm adding a little subscript the t1 and the v1 and the a one is that since we're working on a three -part problem it's very important to specify to differentiate each of the values we're getting because once we go into the second part it's got to be a 2 and v2 and third part is a3 and for a long and sort of complicated physics problems like this it's very important to organize your variables so you know what you're plugging where also it doesn't hurt to do a idiot check every down then okay so yeah looks like we don't have to do any uh rearranging of the function since we're already trying to solve for x1 so all we need to do is just plug everything in so let's go ahead and do that so x initial like we said up here is just going to be 0 plus v initial times t1 as you remember our initial velocity is zero so that means this whole part of the equation will be equal to zero and then for the last part plus one half a t squared.

06:56 It would be one half times our acceleration, which we solved up here.

07:02 So it would be two meters per second squared times times squared.

07:12 As you remember, our time t1 is equal to four seconds.

07:17 So that would be four times four or 16.

07:23 So it would be 16.

07:27 Seconds squared.

07:29 Make sure when you're squaring a variable to also square its units, a good thing to remember.

07:36 So, come on.

07:42 Cool.

07:43 So let's just solve this out.

07:49 So cancel some of these things.

07:52 So one half times two, that'd just be one.

07:54 So these would cancel out.

07:57 Yeah, then we're just left with 16.

08:00 And let's also look at the units two.

08:02 We have meters up here second squared down here then second squared up here these second squares would cancel out so we're just left with meters so this would just be equal to 16 meters and like i said before it's important to also double check the units too we're solving for uh distance distance of course is measured in meters so it makes sense that we're getting meters in the final answer and yeah that looks to be about right okay so that's that for the second part the oh i apologize i realized that i missed a miss something in the first part of the problem you know i apologize for that so that would be acceleration is the direction of the acceleration upward or downward so that's an important thing to remember too let me move all this down is that even though we got a positive number for the acceleration up here it's important to remember that we are moving downwards in this problem and i'm just keeping these numbers positive just make it easier to do the math but in reality we are moving downwards so the acceleration is two meters per seconds but it's two meters per seconds if you want to flip the signs on these things in order to make it make a bit more sense throughout the problem then you can absolutely do that or just remember that the downward direction is going to be the positive direction you know the positive direction which will be important because also means that the direction of gravity will be positive as well when we end up coming to that.

11:03 So it's a important thing to remember.

11:05 And yeah, the direction of the acceleration is going is going to be downwards.

11:11 Okay.

11:13 Let's see.

11:14 Did i, that's it for the second question, for the third question.

11:20 What is the true weight of the man? so true weight, let's first do three.

11:31 So weight, of course, would be, is easy.

11:33 Equal to the mass of the object times the gravitational force, which of course is just going to be 9 .8 meters per second squared.

11:48 What's the saying with the true weight is the weight of the man irrelevant of everything else.

12:00 So the weight would just be plugging in the mass.

12:03 The mass is 50 kilograms.

12:07 So that would be 50 kilograms times 9 .8 meters per second squared.

12:18 I suck at mental math.

12:20 Let me plug it in real quick.

12:22 That'd be 50 times 9 .8.

12:26 And then we get equal to 490 newons or 490 kilograms.

12:46 Meters per second squared.

12:48 Well, let's just answer in newtons to be, to summarize it.

12:55 But yeah, that's the mass.

12:56 No matter what's happening to this man, his true weight will always be 490 newtons, no matter how fast he's traveling or anything.

13:14 True weight, i should say.

13:16 So yeah, that should be for that problem.

13:20 So for four, it's asking what is the net force acting on the man.

13:41 It's the true way the man was the net force acting in to produce the acceleration.

13:45 Okay.

13:46 So, yes, it's asking for the net force required to create this acceleration.

13:56 Again, so important to remember, the equation for force is, is, fnet is equal to mass times acceleration and this case we're using acceleration one so all we need to do is basically just plug everything in we already have each part of this function so that would be equal to 50 kilogram times the acceleration which to triple check is two meters per second squared.

14:54 So that would be two meters per second squared, which would end up being 100 fig newons.

15:11 And again, this is the force of the acceleration going down.

15:16 This is not the weight force.

15:20 This would be the weight force plus some other secondary force.

15:27 Should this be the force on his feet pushing upwards in order to slow him down? okay.

15:38 So that's that for question four.

15:42 Let's see.

15:44 Five was the force exerted on the man's feet by the floor of the elevator? okay.

15:49 So like i was saying.

15:53 So if we draw a simple diagram, let's say that this box is a man.

16:05 He's a very boxy man.

16:06 Then he is on the spring scale.

16:14 So this box is weighs 50 kilograms.

16:22 And so it will have a weight force going downwards.

16:27 That will be the weight force.

16:29 But of course, this box isn't just free falling.

16:32 So it must also have a force pushing up on it, which will be the normal force.

16:42 Of the spring scale pushing up onto the box um yeah of the spring scale pushing up onto the box so what this problem is asking us is to find this normal force and if we make a quick force diagram yes so if not and also i'm putting on this little directional arrow on the forces because right now since we have two forces opposing one another it's important to a specify the directions.

17:29 Whereas with these questions, it really didn't matter.

17:32 We're just finding the absolute value of the forces.

17:38 Just to clarify.

17:40 But for this problem, it's going to be f net is equal to the weight force plus the normal force.

17:52 And like i said above, we're treating the downwards direction to be positive.

17:57 So the normal force that's pointing up is going to be a negative force.

18:03 So and the nice thing is that we have the values for f -net and the weight force.

18:12 So we can just plug those in.

18:16 So that would be 100 newtons is equal to what was it? 490 newtons plus the normal force.

18:34 Again we're solving for the normal force.

18:37 So all we need to do is just subtract 400, 490 newtons on each side and that will get us the normal force which is negative 390 newtons okay so again since this force is pointing upwards the direction of the force is going to be negative the absolute value of the force would just be 390 newtons okay and yeah of course the normal force is the force acting on the man's feet pushing upwards so that should be the fifth part of the problem so the next part what is the apparent weight of the man in newtons that is problem number six so what is the apparent weight of the man in newtons so the apparent weight, that would be, that would basically be the weight pushing on the spring scale.

20:27 Because if you can imagine, if you're trying to weigh yourself in an elevator, your apparent weight would be the number, the scale would give you.

20:38 Which would be the weight force.

20:47 I would think that the apparent weight of the man would just be, it would be the normal force.

20:58 Yeah because of course remember the one of the equations of one newton's equation or one newton's laws is that for every force there's an opposite in equal force or the force in an opposite direction of equal magnitude so if the man is standing on the spring scale and the spring scale is pushing up on the man, that would be the normal force of the man...

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