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 .
2)
Find the total change in its potential energy.
3)
Find the energy dissipated by friction.
4)
Find the power output of the train’s engines.

asked Nov 12, 2014 in PHYSICS

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× 10J

KE = 22.5  MJ

The kinetic energy of the train is 22.5  MJ

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 × 10J

The total change in potential energy is 12348 × 10J

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

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× 10J

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 × 10J

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