Question

a 200 kg roller coaster starts from rest. the ride reaches a maximum height of 30 m and a top speed of 30 m/s. 40 seconds after starting, the roller coaster comes to rest on a section of track that is 5.0 m above the starting height. there are no elastic (spring) forces involved. for the whole ride, the total work done on the roller coaster is _____ and the work done by gravity is _____.

Answer 1

Answe

given,

mass of roller coaster = 200 Kg

maximum height reached = 30 m/s

top speed = 30 m/s

time = 40 s

no elastic spring force is acting so, the total energy is conserved in the system.

The potential and kinetic energy of the roller coaster is conserved in this system.

We know that work done by the conservative force is equal to here and in this case there is no non conservative force like frictional force or spring force acting on it.

Hence work done by the roller coaster will be zero.

a) Work done

`W = KE_2 - KE_1`

`W =(1)/(2)mv^2 - (1)/(2)mv^2`

`W =0\ J`

b) Work done by gravity

W = m g h

W = 200 x (-9.8 ) x 5

negative sign represent work done by the gravitational force is in negative direction.

W = -9800 N

Answer 2

Total work done is W = 0 j

work done by gravity ,W = - 30000 J ON THE BASIS OF TOP SPEED

Step-by-step explanation:

given data:

mass of roller coaster is 200 kg

max height is 30 m

top speed is 30 m/s

height of section from starting point is 5 m

from work energy theorem

Work done
`= \Delta U_(G) + \Delta U_(E) + \Delta KE`

as there is no elastic force so we have

work done is
`\Delta U_(G) + \Delta KE`

total work done is
`W = 0 + \Delta KE`

`W = (1)/(2) mv^2 + (1)/(2) mv^2`

W = 0 j

work done by gravitational force is W = mgh

`W = 200 * 30 * 5 = 30000 J`

W = - 30000 J ON THE BASIS OF TOP SPEED

here -ve sign indicate that body moves in opposite direction of gravity force