Physics help please!

Question

1)

A train with a total mass of 1.8 × 106 kg rises 
700 m while traveling a distance of 53 km at 
an average speed of 18 km/h. The frictional 
force is 0.8 percent of the weight. 
Find the kinetic energy of the train. The 
acceleration of gravity is 9.81 m/s2 . 
Answer in units of MJ 
2)
Find the total change in its potential energy. 
Answer in units of J 
3)
Find the energy dissipated by friction. 
Answer in units of J 
4)
Find the power output of the train’s engines. 
Answer in units of MW

Answer 1

1)

Train total mass m =1.8× 10⁶ kg

Rising height h = 700 m

Traveling distance d = 53 km

d = 53000 m

Average speed v = 18 km/h

v = 18*(5/18) m/s

v = 5 m/s

The acceleration of gravity g = 9.81 m/s²

The kinetic energy of the train KE = ?

KE = ½mv²

KE = ½(1.8× 10⁶)(5)²

KE = 22.5× 10⁶  J

KE = 22.5  MJ

The kinetic energy of the train is 22.5  MJ

Answer 2

2)

Train total mass m =1.8× 10⁶ kg

Rising height h = 700 m

Traveling distance d = 53 km

d = 53000 m

Average speed v = 18 km/h

v = 18*(5/18) m/s

v = 5 m/s

The acceleration of gravity g = 9.81 m/s²

At starting height is h1 = 0 m

At ending height is h2 = 700 m

Change in height Δh = h2 - h1

Δh = 700 - 0

Δh = 700 m

The total change in potential energy ΔPE = ?

ΔPE = mgΔh

ΔPE = (1.8× 10⁶)(9.8)(700)

ΔPE = 12348 × 10⁶  J

The total change in potential energy is 12348 × 10⁶  J

Answer 3

3)

Train total mass m =1.8× 10⁶ kg

Rising height h = 700 m

Traveling distance d = 53 km

d = 53000 m

Average speed v = 18 km/h

v = 18*(5/18) m/s

v = 5 m/s

The acceleration of gravity g = 9.81 m/s²

The frictional force is 0.8 percent of the weight

Ff = ( 0.8 / 100 )(Weight)

Ff = ( 0.8 / 100 )mg

Ff = ( 0.8 / 100 )*1.8*10⁶*9.8

Ff = 141120

The energy dissipated by friction Ef = ?

The energy dissipated Ef = Frictional force x distance

Ef = Ff x d

Ef = 141120 x 53000

Ef = 7479360000

Ef = 747936 × 10⁴  J

The energy dissipated by friction is 747936 × 10⁴  J

Answer 4

4)

Train total mass m =1.8× 10⁶ kg

Rising height h = 700 m

Traveling distance d = 53 km

d = 53000 m

Average speed v = 18 km/h

v = 18*(5/18) m/s

v = 5 m/s

The acceleration of gravity g = 9.81 m/s²

The frictional force is 0.8 percent of the weight

Ff = ( 0.8 / 100 ) mg

Ff = ( 0.8 / 100 )*1.8*10⁶* 9.8

Ff = 141120

The power output of the train Pt = ?

The kinetic energy of the train KE

KE = ½mv²

KE = ½(1.8× 10⁶)(5)²

KE = 22.5× 10⁶  J

At starting height is h1 = 0 m

At ending height is h2 = 700 m

Change in height  Δh = h2 - h1

Δh = 700 - 0

Δh = 700 m

The total change in potential energy ΔPE = mgΔh

ΔPE = (1.8× 10⁶)(9.8)(700)

ΔPE = 12348 × 10⁶  J

The energy dissipated due to friction Ef = Frictional force x distance

Ef = Ff x d

Ef = 141120 x 53000

Ef = 7479360000

Ef = 7479.36 × 10⁶  J

Total energy Et = KE+PE+Ef

Substitute KE = 22.5× 10⁶ , PE = 12348 × 10⁶ , Ef = 7479.36 × 10⁶

Et = (22.5× 10⁶)+(12348 × 10⁶)+(7479.36× 10⁶)

Et = 19849.86 × 10⁶  J

Total force = Total energy / distance

Ft = Et / d

Ft = (19849.86 × 10⁶ ) / 53000

Ft = 0.37452566 × 10⁶ N

Total power Pt = Total Force × velocity

Pt = (Ft)v

Pt = (0.37452566 × 10⁶)(5)

Pt = 1.87263 × 10⁶ W

Pt = 1.87263 MW

The power output of the train is 1.87263 MW