A roller Ouster starts on top of the first hill that is 248 ft high. In moving d

Question

A roller Ouster starts on top of the first hill that is 248 ft high. In moving down hill, 6.0% of the starting potential energy is lost due to friction. What is the velocity coaster at the bottom of this hill? 4. A force of 7.2 N acting on a 1.0-kg object initially at rest moves it through a parallel distance of 2.5 m. a. How much work is done on the object? b. Neglecting friction, what is the speed of the object when it is moved 2.5 m? 5. An 80.0 foot long bridge weighing 600 lbs is stretched between two supports that are 50 feet apart. (See the sketch below.) If two loads of the weights and locations shown in the sketch are placed on the bridge, what are the forces on the two bridge supports at points A and B? The requirements for the elevator at the John Hancock Building in Chicago state that it must travel from the first floor to the eighty-third floor (a height of 850 ft) in 25 seconds. The weight of the empty elevator car is 1,600 lbs, and the maximum weight of its contents (people and equipment) is 2,400 lbs (the total weight is the sum of these weights). What is the smallest electric elevator motor (in horsepower) that will satisfy these requirements?

Explanation / Answer

(3) h= 248ft=75.59m

potential energy=mgh=m*9.8*75.59 joule

6 percent of potential energy is lost due to friction

remaining potential energy=94 percent=m*9.8*75.59*94/100

this is the kinetic energy at the bottom of the hill. because potential energy is zero(because h=0). and energy conservation says total energy remains constant.

(1/2 )mv2 =m*9.8*75.59*94/100
so v2=19.6*75.59*9.4=13926.70

v=118.01m/sec

4(a) work W= FS cos(theta)

F is force S is displacement and theta is zero

W=7.2*2.5*cos0=18 joule

(b) from F=m*a here a is acceleration

a=7.2/1.0=7.2m/sec2

here initial velocity is zero because particle starts from rest.

from equation of motion

v2 =u2+2as

v is final velocity

v2 =0+2*7.2*2.5=36 m/sec

v= 6 m/sec

(5)let the forces on two support A and B are m1g and m2g

(400+500)lbs=600lbs +m1g+m2g

408.23kg=272.16+m1+m2

m1+m2=136.07kg------------------(1)

600lbs=272.16 worked at the center(40 ft from A)

according to rule of sum of moments

m1g*40ft+181.44*20ft=m2g*10ft+226.80*40ft

converting feet to meter

m1g*12.19+181.44*6.1=m2g*3.04+226.80*12.19

m1g*12.19+1106.784=m2g*3.04+2764.69

m1g*12.19-m2g*3.04=1657.91----------(2)

m1*12.19-m2 *3.04 =169.17

putting the value of m1 from eq (1)

(136.07-m2)*12.19-m2*3.04=169.17

1658.69-12.19m2-3.04m2=169.17

1489.52=15.23m2

m2=97.99kg

puttingthis in eq (1)

m1=38.08kg

so force at A and B are

m1g and m2g=373.18N and960.30N

(6) work done is stored in the form of potential energy=mgh

and power =rate of doing work=w/t

m=1600+2400=4000lbs=1814.37kg

h=850ft=259.08m

g=9.8 m/sec2

mgh= 1814.37*259.08*9.8=46.06*105 joule

P= 46.06*105/25=1.84*105 watt=246.65hp

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