5

A 25.0 g mass rests on a 50.09 inclined plane (4k = 0.200). It is pushed against spring, compressing it 15.0 cm; and released. It moves Up the plane and leaves the ...

Question

A 25.0 g mass rests on a 50.09 inclined plane (4k = 0.200). It is pushed against spring, compressing it 15.0 cm; and released. It moves Up the plane and leaves the spring, traveling an additional 45.0 cm before leaving the edge of the plane (at a height of 80.0 cm above the ground) It then flies through the air; and is observed to land 270_ cm from the base of the incline_ What was the spring constant?CM270 cin

A 25.0 g mass rests on a 50.09 inclined plane (4k = 0.200). It is pushed against spring, compressing it 15.0 cm; and released. It moves Up the plane and leaves the spring, traveling an additional 45.0 cm before leaving the edge of the plane (at a height of 80.0 cm above the ground) It then flies through the air; and is observed to land 270_ cm from the base of the incline_ What was the spring constant? CM 270 cin



Answers

An inclined plane of angle $\theta=20.0^{\circ}$ has a spring of force constant $k=500 \mathrm{N} / \mathrm{m}$ fastened securely at the bottom so that the spring is parallel to the surface as shown in Figure P6.61. A block of mass $m=2.50 \mathrm{kg}$ is placed on the plane at a distance $d=0.300 \mathrm{m}$ from the spring. From this position, the block is projected downward toward the spring with speed $v=0.750 \mathrm{m} / \mathrm{s}$. By what distance is the spring compressed when the block momentarily comes to rest?

Hi Friends as solid difficult There is a block of mask 15 Kidding In contact with my spring of the spring Constant 14 new 10 per meter over the in client Mhm Mhm Friction Right Service having inclination Mhm targeted. We have to fire the maximum distance Moved Uh huh. By the block. Mhm! And how much does state it will move while returning Bent would be airplane Mhm Conservation of energy What changing kind of technology Changing potential energy is got to energy lost by those people It's zero mgt! Sign of data minus This is lost half cape X Finally square substitute observing the mass of the block White five kg Do you 9.8 d We are finding this is signed 30 half K is 40 Final group initial is 0.2 square. So distressed moved by the block is Why did she do? We did fish. There is no loss of energy. Yeah. Mm hmm. So during returning them Oh not been moved. Why do you to eat it? That sucks

Everyone, This is the problem is gone. Contribution off energy as soul In the figure there is inclined plane having inclination 20 degree at the bottom. Off it I spring off in constant 500 Newton permit er is fixed and at the top of it there is a block off must 2.5 KD at at the state's off while 3 m from this big it is thrown towards the spring with the speed off 1 75 meter per second. We have to fire compression industry not to see it. We have to apply contribution off energy according to contribution off energy A knish ALS kind of technology Initial gravitational potential energy initial spring energy Yeah, it's called toe Final Kind of technology Final gravitational potential energy Final spring energy No substitute the value and each other kind of technology. Half M v I script gravitational potential energy you can find by and D H Final kind of technology zero because it is coming Moment Lee addressed is initial spring energy Europe because it is unexpressed. You know, gravitational potential energy becomes zero on it is half x squared. Uh huh. Half and yeah, Mhm. Yeah. Edge will be deep Bliss. Hex. Sign off Ito. Mm. Into G edge. Just a moment. No half and re I squared. Plus MGH having this value is there. Toe half gives square. No substituting the value. Uh huh. Mask off. The body is given to 25 Any silver cities? Why am 75 ever? 2.5 and two G 9.8 D is given three 13 m. Sign off 20. Yeah, it's going toe half cage 500. Mhm. And to access for solving this value for eggs, they will get Yeah, 0.135 After solving this quality execution value off actually will get 13 point meter, that's all. Thank.

For this problem on the topic of energy conservation, we're told that a 0.5 kg mass is propelled up an incline by the use of a spring with spring constant 500 newtons per meter. The spring is initially compressed, but that he centimeters from its equilibrium position and sends the mass from rest firstly across a horizontal surface and then onto the plane. The plane has a length of four m an incline of 30° to the horizontal. The coefficient of kinetic friction between the mass and the surface is 0.35. And When compressed this, when the spring is compressed, the masses 1.5 m from the bottom of the plane. We want to find the speed of the mass as it reaches the bottom of the plane. The speed of the mass as it reached the top of the plane. And the total work by friction from the beginning to the end of the masses motion. Now the elastic potential energy is you spring. Yeah. And this is equal to half K X squared where K is the spring constant? And X is the compression. The mass loses energy which is the work done by friction, which is minus the frictional force, F F times the distance that the mass traveled. D. And this is equal to minus UK times M G. D. Due to friction. And therefore the kinetic energy at the bottom is given by K B. And K B is equal to a half M. The B squared which is equal to half K X squared minus the work done due to friction, Mieux, que times am times G times D. And so from here we can rearrange to find the speed of the block at the bottom of the plane. VB to be the square root of K X squared over two K X squared over em rather minus two UK jean times deep. And so since all these values are known if we substitute them in, we get this to be the square root of 500 newtons per meter times the compression of the spring, 0.3 m squared divided by The mass of the block, 0.5 Kg -2 times. The coefficient of kinetic friction is 0.35 Times the acceleration due to gravity 9.81 m/km2 Times a distance of 1.5 m which gives us the speed of the block at the bottom To be eight 93 meters per second. So that's the block speed just before it starts moving up the inclined plane. Now, for part B we have that to reach the top of the incline. The gravitational potential energy must also be considered. And the change of gravitational potential energy, delta U. G. Is equal to the potential energy at the top minus the potential energy at the bottom. Now, since the plane has lent l the an incline angle theater, the change in gravitational potential energy is MGl scientist to. And the kinetic energy at the top can then be calculated by subtracting the gravitational potential energy and work due to friction from the kinetic energy at the bottom. So Kay top is equal to okay at the bottom minus UK MG L. Sign theater minus mgl sine theta. And so by rearranging this equation, since we know half um the top squared is equal to K B minus UK mgl signed data minus MG al sine theta. We can find the speed of the block at the top of the incline, the top to be the square root of two over. Em into the kinetic energy at the bottom minus M G L. Into UK. Call sign data plus sign data. And so if we substitute the values into this equation, we get this to be the square root of two over 0.5 Kg. And the kinetic energy at the bottom. We can get from VB, which we calculated in part A and it's a half M V B squared which is 19 0.92 jules, yeah -0.5 Kg In 29.81 m/km2 times for meters multiplied by 0.35 Times The co sign of 30° plus the sign of 30 degrees. And so if we calculate this, we get the speed at the block at the top To be 4.08 m per second. And lastly, we want to find the total work done by friction from the beginning to the end of the motion of the mass. Now the total work done due to friction is equal to minus the frictional force ff in two D plus L, which is minus UK m g d minus UK mg. Call sign of theater times L. And so if we substitute our values into this equation, this is minus 0.35 times zero 0.5 kg Times 9.81 m/km2 Into 1.5 m less, four m times the call sign of 30 degrees. And so this gives us the work done due to friction for the blocks, entire motion to B -8 0.52 jules.

Uh huh.


Similar Solved Questions

5 answers
Homework: Math Fitness: Training Challenge | Opt Score: 0 of 13 of 27 (25 complete 3.4.21Solve.2x+ 6complote Your choice_ choice below and_ necessary; fill in the answer box Select the correctThe solution(s) islare separate answers as needed ) simplified fraction Use comma (Type an integer Or = {xIxis = rea number and x# The solution set is comma separate answers as needed: ) simplified fraction. Use (Type an integer Or There no solution_
Homework: Math Fitness: Training Challenge | Opt Score: 0 of 13 of 27 (25 complete 3.4.21 Solve. 2x+ 6 complote Your choice_ choice below and_ necessary; fill in the answer box Select the correct The solution(s) islare separate answers as needed ) simplified fraction Use comma (Type an integer Or =...
5 answers
3-12 List the first five terms of the sequence (-1)"-= 5. Un 5"
3-12 List the first five terms of the sequence (-1)"-= 5. Un 5"...
5 answers
22.[35coledIhrs JnsctcoleclProblem 2. FoiliLel IncientlIenaihi ul 4 diilacni recinor wloxudc kuths Cotailnt Tat J40A ULceasc_ consal Fill: Diand Tiunit tha vary Duh nn asle the sue ol Iha deJofal changing AtenTrv=FAnswciEntered Anttrt PreyledRezulinconeccorecCneettalteregneent ,
22.[35 coled Ihrs Jnsct colecl Problem 2. FoiliLel Incientl Ienaihi ul 4 diilacni recinor wloxudc kuths Cotailnt Tat J40A ULceasc_ consal Fill: Di and Tiunit tha vary Duh nn asle the sue ol Iha deJofal changing Aten Trv=F Answci Entered Anttrt Preyled Rezuli nconec corec Cneettaltereg ne ent ,...
5 answers
Consider the solubilitics of particular solule at two diflerent temperaturesTeniperature ( C) 20.0Sclubility (g/1CO g H;O) 37.2 85.130.0saluraled solution of this solute Wus made using 720 g H,O # 20.0 C, How much more solute can bc added Suppost if the lemperiure is increused t0 300 "C?mns104
Consider the solubilitics of particular solule at two diflerent temperatures Teniperature ( C) 20.0 Sclubility (g/1CO g H;O) 37.2 85.1 30.0 saluraled solution of this solute Wus made using 720 g H,O # 20.0 C, How much more solute can bc added Suppost if the lemperiure is increused t0 300 "C? mn...
5 answers
RequiredWhat is the net ionic equation of the following reaction? HNO: (aq) + KOH (aq) KNO: (aq) + HzO (L)A Ht + NOs + Kt + OH _ Kt+NOs + Ht + OH B. H + NOs + Kt + OH _ K-+NO; + HzO C. H + OH HzO D. Kt + NO; KNOz
Required What is the net ionic equation of the following reaction? HNO: (aq) + KOH (aq) KNO: (aq) + HzO (L) A Ht + NOs + Kt + OH _ Kt+NOs + Ht + OH B. H + NOs + Kt + OH _ K-+NO; + HzO C. H + OH HzO D. Kt + NO; KNOz...
5 answers
The length of timc; in bour8, It taks group of pcople; 40 ycars and older, to play one soccer match is pormally distributed with mean of 2 bours and standard deviation 0f 0.5 hourS: sample of gize 50 js drawn randomly from the population _ Find thbe probability that the samplc mcan is less than 2.3 hours:Write the probability in terms of thc statistic Write the probability in terms of the %-scOrCWrite tbe probability in numeric , decimal form:
The length of timc; in bour8, It taks group of pcople; 40 ycars and older, to play one soccer match is pormally distributed with mean of 2 bours and standard deviation 0f 0.5 hourS: sample of gize 50 js drawn randomly from the population _ Find thbe probability that the samplc mcan is less than 2.3 ...
5 answers
Point) Consider the following initial value problem:y" _ Ty' 18y sin(3t)y(0) = 5, % (0) = 5Using Y for the Laplace transform of y(t) ie, Y = C{ylt)}, find the equation you get by taking the Laplace transiorm of the differentia equation and solve for Y(s)
point) Consider the following initial value problem: y" _ Ty' 18y sin(3t) y(0) = 5, % (0) = 5 Using Y for the Laplace transform of y(t) ie, Y = C{ylt)}, find the equation you get by taking the Laplace transiorm of the differentia equation and solve for Y(s)...
5 answers
Te iirst term of a geometric sequence Is 25, and the fourth term Is(a) Find the common ratloFind the fifth term a5(6) Find the partial sum of the first eight terms 58
Te iirst term of a geometric sequence Is 25, and the fourth term Is (a) Find the common ratlo Find the fifth term a5 (6) Find the partial sum of the first eight terms 58...
3 answers
4. Consider the following R output, which summarizes weights from the databank data set, by gender:gender 50 50mean 129 . 170 38sd 20 . 21 23.37Minimum Q1 Median Maximum 99 112 125 . 01 143 179 .0 128 00 152 00 167 00 187 25 234 00Fill in the blank Approximately weigh at least 112 pounds? (a) 13 (b) 38 (c) 75 (d) 25 (e) 43(how many?) females in the databank data set
4. Consider the following R output, which summarizes weights from the databank data set, by gender: gender 50 50 mean 129 . 170 38 sd 20 . 21 23.37 Minimum Q1 Median Maximum 99 112 125 . 01 143 179 .0 128 00 152 00 167 00 187 25 234 00 Fill in the blank Approximately weigh at least 112 pounds? (a) 1...
1 answers
Rework Problem 103 if the reduction is to be $20 \%$.
Rework Problem 103 if the reduction is to be $20 \%$....
1 answers
Solve each system. State whether it is an inconsistent system or has infinitely many solutions. If the system has infinitely many solutions, write the solution set with z arbitrary. See Examples $3,4,6,$ and 7 $$\begin{aligned} &5 x-4 y+z=0\\ &x+y=0\\ &-10 x+8 y-2 z=0 \end{aligned}$$
Solve each system. State whether it is an inconsistent system or has infinitely many solutions. If the system has infinitely many solutions, write the solution set with z arbitrary. See Examples $3,4,6,$ and 7 $$\begin{aligned} &5 x-4 y+z=0\\ &x+y=0\\ &-10 x+8 y-2 z=0 \end{aligned}$$...
5 answers
Write the name for each of the following.CH4CH3(CH2)7CH3CH3CH3CH3(CH2)5CH3CH3CH2CH3CH3(CH2)3CH3CH3(CH2)2CH3CH3(CH2)8CH3CH3(CH2)4CH3CH3(CH2)6CH3
Write the name for each of the following. CH4 CH3(CH2)7CH3 CH3CH3 CH3(CH2)5CH3 CH3CH2CH3 CH3(CH2)3CH3 CH3(CH2)2CH3 CH3(CH2)8CH3 CH3(CH2)4CH3 CH3(CH2)6CH3...
5 answers
A2.0 kg mass lies along the y-direction distance of 8.0 m from the origin, and 5.0 kg mass lies along the x-direction distance of -3.0 m from the origin: Determine the distance from the origin (in m) t0 the center of mass,
A2.0 kg mass lies along the y-direction distance of 8.0 m from the origin, and 5.0 kg mass lies along the x-direction distance of -3.0 m from the origin: Determine the distance from the origin (in m) t0 the center of mass,...
5 answers
572 1 1 JL 1 1
572 1 1 JL 1 1...
5 answers
Mtlo (yl*o 0(t) k ~[s)F(s) &s slves Ute Initia probien F(t} x(tul above the @ ven Intlal-valve problamwhanever @(r
mtlo (yl*o 0(t) k ~[s)F(s) &s slves Ute Initia probien F(t} x(tul above the @ ven Intlal-valve problam whanever @(r...
5 answers
(8 points) Find the composition,9, using the functions below_ *+5 if* <-10 f(x) = X2r~ 1 if -10 <x<0 3x + 2 if*20g(r) = (-29 "{{128 #f * <0
(8 points) Find the composition, 9, using the functions below_ *+5 if* <-10 f(x) = X2r~ 1 if -10 <x<0 3x + 2 if*20 g(r) = (-29 "{{128 #f * <0...

-- 0.021177--