resistance work-done speed

Question

3.2. a trolley with a mass of 310 kg starts to move downward from rest of an incline of 17 degrees with the horizontal. The tractive resistance is 125N and the incline is 22m long. Calculate the speed of the trolley at the end of the incline 3.3. a truck with a mass of 84 tons is pulled 1070,8 m up a 1:0,04 incline at a constant speed of 74 km/h. Calculate the following: 3.3.1. the work done against gravity, in MJ 3.3.2. the total work done in MJ, if the tractive resistance is 75 N/ton.

Answer 1

3.3)

Mass of the truck is  84 tones  =  84000 kgs 

Distance travelled by truck 1070.8 m

Slope  expressed as a ratio 1:0.04 .

To convert slope percentage to degrees we use formula  ø =  tan^-1 (1 / n) 

 ø =  tan^-1 (1 /0.04) 

 ø =  tan^-1 (25)

 ø = 87.709°

So the Ramp is inclined with  87.709° .

3.3.1)

The force against the gravity is mg sin ø .

                                                    = 84000*9.8 sin 87.709

                                                    = 823200*0.9992

                                                   = 822541.44 N .

So the force against the gravity is 822541.44 N .

The work done against the gravity is = force * distance

                                                    = 822541.44*1070.8 

                                                    = 880777374 J

                                                    = 880.77 MJ .

The work done against the gravity is  880.77 MJ .

Answer 2

3.3.2)

Mass of the truck is  84 tones  =  84000 kgs 

Distance travelled by truck 1070.8 m

Slope  expressed as a ratio 1:0.04 .

To convert slope percentage to degrees we use formula  ø =  tan^-1 (1 / n) 

 ø =  tan^-1 (1 /0.04) 

 ø =  tan^-1 (25)

 ø = 87.709°

So the Ramp is inclined with  87.709° .

The force against the gravity is mg sin ø .

                                                    = 84000*9.8 sin 87.709

                                                    = 823200*0.9992

                                                   = 822541.44 N .

So the force against the gravity is 822541.44 N .

The work done against the gravity is = force * distance

                                                    = 822541.44*1070.8 

                                                    = 880777374 J

                                                    = 880.77 MJ .

The work done against the gravity is 880.77 MJ  .

Given the tractive resistance is 75 N/ton .

So the  tractive resistance for 84 tones  is 75*84 = 6300 N

So the work done for  tractive resistance is = force * distance

                                                               = 6300 * 1070.8

                                                              = 6746040 J

                                                             = 6.746 MJ

So th total work done = (work done against the gravity) + (work done for  tractive resistance)

                              = 880.77 + 6.746

                             = 887.516 MJ

The total work done is 887.516 MJ .

Answer 3

3.2)

Mass of the trolley is  310 kg

Distance traveled by trolley 22 m

The Ramp is inclined with an angle of 17° .

The tractive resistance is 125 N .

The total force = (force against the gravity) + (force for  tractive resistance)

                   F = mg sin ø   + 125 

                   F = (310)9.8 sin 17   + 125   

                   F = 888.225   + 125  

                   F = 1013.225  N .

So the total force is given by 1013.225  N .

Acceleration of the trolley can be found using a = F/m                            [ Since F = ma ]

a = 1013.225 / 310 

a = 3.26 m/s²

So the acceleration of the trolley is 3.26 m/s² .

The trolley starts from rest so the initial velocity is zero . 

The velocity of the trolley can be found using formula V_f² - V_i²  = 2as .

V_f² - 0² = 2(3.26)(22) 

V_f² = 158.4 

Vf = 12.58 m/s .

So the velocity /speed of the trolley is 12.58 m/s .