Welcome :: Homework Help and Answers :: Mathskey.com
Welcome to Mathskey.com Question & Answers Community. Ask any math/science homework question and receive answers from other members of the community.

13,435 questions

17,804 answers

1,438 comments

157,629 users

Physics help please!

0 votes

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

asked Nov 12, 2014 in PHYSICS by anonymous

4 Answers

0 votes

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

answered Nov 12, 2014 by Shalom Scholar
0 votes

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

answered Nov 12, 2014 by Shalom Scholar
0 votes

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

answered Nov 12, 2014 by Shalom Scholar
0 votes

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

answered Nov 12, 2014 by Shalom Scholar

Related questions

asked Jan 26, 2016 in PHYSICS by anonymous
asked Nov 29, 2015 in PHYSICS by anonymous
asked Nov 28, 2015 in PHYSICS by anonymous
asked Nov 7, 2015 in PHYSICS by anonymous
...