FINAL PROJECT

RESULTS:

A car with a mass of 950 kg is on top of a hill with a height of 50 m and is at rest.

1. Calculate the potential energy when it is at the top of the hill.

2. The speed when descending the hill ends

3. His speed at 27m height .

1. U=mgh

U=(950 kg) (9.8 m/s2) (50 m)

U= 465500 Joules

K.E= (m.v2)/2

K.E= ( 950 Kg x 0 m/s) /2

K.E= O Joules

E=K.E+U

E= 0 Joules + 465500 JOULES

E= 465500 JOULES

2. U=mgh

U=(950 Kg) (9.8 m/s2) (0 m)

U= 0 joules

E= K.E + U

465500 J= K.E + 0 Joules

K.E= 465500 Joules

K.E=(m.v2)/2

465500 J= (950 kg x v2) /2

931000 J= 950 v2

980 J/kg =v2

31.3 m/s = v

3. U= mgh

U= (950 kg) (9.8 m/s2) ( 27 m)

U= 251370 Joules

E= K.E+U

465500 Joules= K.E + 251370 Joules

K.E= 214130 Joules

K.E= (m.v2)/2

214130 J= (950 kg x v2) / 2

428260 J= 950kg v2

428260 J/Kg= 950 v2

654.42 m/s = v

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