1. A COMPACT CAR HAS A MASS OF 950 KG AND AN OVERALL EFFICIENCY OF 23.0% (THAT I

Question

1. A COMPACT CAR HAS A MASS OF 950 KG AND AN OVERALL EFFICIENCY OF 23.0% (THAT IS IT CONVERTS 23.0% OF CHEMICAL ENERGY STORED IN THE GASOLINE INTO WORK DONE BY THE DRIVE TIRES PUSHING ON THE ROAD)

(FOR 1B AND 1C ASSUME CONSTANT ACCELERATION OF 2.00 M/S^2)

A) IF ONE GALLON OF GASOLINE CONTAINS 1.34*10^(8) JOULES OF STORED CHEMICAL ENERGY, FIND THE AMOUNT OF GASOLINE (IN GALLONS) USED WHEN THE CAR ACCELRATES FROM REST TO 32 M/S (WITH NO AIR OR ROAD FRICTION AND ON THE LEVEL).

B) REPEAT PART A FOR THE CASE WHERE THE UNIFORM ACCELERATION AND AN AVERAGE FRICITIONAL FORCE TOTAL AIR AND ROAD OF 370 N

C) REPEAT PART B FOR THE CASE WHERE THE UNIFORM ACCELERATION IS UP A 5.00% SLOPE.

D) THE MILEAGE CLAIMED FOR THE CAR IS 60.8 KM/GALLON AT A SPEED OF 32.0 M/S. WHAT POWER IS DELIVERED TO THE WHEELS( TO OVERCOME FRICTIONAL EFFECTS) WHEN THE CAR IS DRIVING AT 32.0M/S ON A LEVEL ROAD AND WHAT IS THE AVERAGE FRICTIONAL FORCE AT THIS SPEED ON THIS CAR?

2. A SMALL MASS M RESTS AT THE TOP OF A SMOOTH (FRICTIONLESS) HEMISHPERE OF RADIUS R. IT IS DISPLACED SLIGHTLY SO THAT IT SLIDES FROM REST DOWN THE HEMISPHERE. IT REMAINS IN CONTACT WITH THE SURFACE UNTIL IT REACHES THE ANGLE (theta) . FROM THIS POINT ON IT LOSES CONTACT WITH THE SURFACE AND FOLLOWS A FREE-FALL PARABOLIC PATH. FIND THE ANGEL (theta). HINT: YOU MUST USE BOTH CENTRIPETAL FORCE ANALYSIS AND ENERGY ANALYSIS TO SOLVE THIS PROBLEM. AT THE POINT WHERE M LEAVES THE SPHERE, THE NORMAL FORCE IS ZERO.

Explanation / Answer

A)Let u be intial speed = 0 m/s

v be final speed = 32 m/s

Work done = change in Kinetic Energy = 1/2*M*v^2 - 1/2*M*u^2

Work done = 1/2*950*32^2 = 486.4 KJ

Efficiency = Work done/chemicl Energy of gasoline = 0.23

chemicl Energy of gasoline = Work done/0.23 = 486.4/0.23 = 2114.78 KJ = 2.12*10^6 J

1 gallon = 1.34*10^8 J

Therefore

chemicl Energy of gasoline = 2.12*10^6/( 1.34*10^8) gallons = 0.016 gallons

B) v^2-u^2 = 2*a*S

32^2 = 2*2*S

S = 256 m

Work done -Friction force = Change in Kinetic energy

Work done = 370*256 + 486400 Joules

Work done = 5.81*10^5 Joules

chemicl Energy of gasoline = 5.81*10^5/0.23= 2.53*10^6 Joules

chemicl Energy of gasoline = 2.53*10^6/( 1.34*10^8) = 0.019 gallons

C)Tan(angle) = 5/100

Frictional Force = 370*S*Sin(angle) = 370*256*5/100.13 = 4730.1 J

Work done = 4730.1 + 486400 = 4.91*10^5 J

chemicl Energy of gasoline = 4.91*10^5/0.23 = 2.14*10^6 J

chemicl Energy of gasoline = 2.14*10^6/( 1.34*10^8) = 0.016 gallons

D) Work done = 1.34*10^8*0.23 = 3.1*10^7 Joules

S = 60.8 KM

Power delivered = (32/60.8*10^3)*1.34*10^8 Watts = 70.53 KW

Friction*60.8*10^3 = 3.1*10^7-486400 = 3.1*10^7-0.49*10^7 = 2.61*10^7

Friction Force = 429.28 N

2)

First of all, the mass does not lose contact with the surface of the sphere at all, considering it is undergoing centripetal acceleration, where the centripetal force is provided by the component of the mass's weight towards the centre of the hemisphere (mgcos?, where ? is the angle between the vector of the mass's weight and its component acting towards the centre).
Therefore, since only mgcos? is responsible for the centripetal force, I can form a relationship like this:

mgcos? = mv2r

v = rgcos???????

Taking 'h' as the height of the mass from the base of the hemisphere.

cos? = hr-----------(1)

Then the velocity of the mass becomes:

v = gh???

The component of the mass's weight along the centre disappears only when ? becomes 90 degrees. At this point, it leaves the surface of the hemisphere.
Now, the energy of the mass at the topmost point is:

P.E = mgr

As the body slides over the hemisphere's surface, it has a tangential velocity given by the expression I had just previously derived. So by the conservation of total mechanical energy of the body, its energy at any other point on the hemisphere is:

T.E = mgh+12mv2

P.E = T.E

gr = gh+v22

gr = gh+gh2

h = 23r

From !

Cos(Theta) = 2/3

Theta = 48.19 deg

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