POST 2

question 1

A lorry haning a mass of 1.5t is travel-ling along a level road at 72 km/h. when the brakes are applied, the speed decreases to 18 km/h. determine how much the kinetic enegy of the lorry is reduced.

Initial velocity of lorry, v1                  = 72 km/h

                                       km                    m            1h

                            =  72   −−−   x  1000     −−−    x    −−−

                                        H                     km         3600s

                                    72

                               =  −−−   = 20 m/s

                                   3.6

Final velocity of lorry, v2                =  18

                                                        −−−  =   5 m/s and

                                                         3.6

Mass of lorry , m                             = 1.5t = 1500 kg

Initial kinetic energy of the lorry        1                    1

                                                  = −−− mv ²   =   −−−  (1500)(20) ²

                                                      2          ¹        2                   

                                              

             = 300 kj

Final kinetic energy of the lorry         1                  1

                                                 =  −−  mv²  =   −− ( 1500)(5) ²

                                                      2     ²           2

                                                    

 = 18.75 kj

Hence, the change in kinetic energy    = 300 – 18.75

                            

                                                        = 281.25 kj

POST 2

A rod of mass  M=3KG  and length  L=1.2m  pivots about an axis, perpendicular to its length, which passes through one of its ends. What is the moment of inertia of the rod? Given that the rod's instantaneous angular velocity is  60deg/s  what is its rotational kinetic energy?

Answer: The moment of inertia of a rod of mass M and length L about an axis, perpendicular to its length, which passes through its midpoint is I = (1/12)ML². This is a standard result. Using the parallel axis theorem, the moment of inertia about a parallel axis passing through one of the ends of the rod is

I¹ I + M ( L / 2) ²=1/3 ML²

I¹ = 3 x 1.2² =1.44 kg m²

         −−−−−

            3

                                 

The instantaneous angular velocity of the rod is 

w = 60 x  π  = 1.047 rad/s

           −−−

            180

Hence, the rod's rotational kinetic energy is written 

K = 1 I¹ w² = 0.5 x 1.44 x 1.047² = 0.789 J

        −

        2

APPLICATION

Question 1.

How does the movement of the earth affects the distribution of solar energy in the earth's surface?

Solution: These two phenomena of circulation of the earth and the earth's rotation on its  axis is the variation of solar influence on the earth surface. he earth's rotation  axis tilt of the angle 231 / 2 degrees is causing the phenomenon of day and  night occur. Earth takes 24 hours to make one complete rotation. The Earth  rotates west to east. The surface of the earth facing the sun during the day and  would have received a lot of solar energy distribution compared with that  experienced on earth during the night.

                            Question 2.

How rotation engine work?

Solution 

Rotation engine is a type of internal combustion engines that use design to change
 the rotation of the pressure of combustion air-fuelmixture directly to the rotational motion,
 unlike ordinary piston engine which produces a reciprocating motion before
 being transferred to the rotational movement of the crankshaft. In the four-stroke cycle
 engine generated in the space between the housing shaped like oval epitrokoid
 the triangular rotor. This design produces a smooth high-rpm power from a compact
 and lightweight engine.

GROUP NAME

• 
  

          GROUP:ENGSCIENSE

1) Amirul Akhmar B Mohd Sabri

2) Mohammad Anuar Fitri B Sahimi

3) Mohammad Shafiqal Bin Mat Saad

4) Mohammad Syafiq Fitri Bin Shaarani