The roller-coaster car starts its journey from a vertical height of 40 m on the first hill and reaches a vertical height of only 25 m on the second hill, where it slows to a momentary stop. It traveled a distance of 400 m. Determine the thermal energy produced and estimate the average friction force on the car whose mass, is 1000 kg. (Ans. 147,150 J, 370 N)


Given:

Height of the first hill = h1 = 40 m
Height of the second hill = h2 = 25 m
Total distance travelled = d = 400 m
Mas of the car = m = 1000 kg
Value of g = 9.81 m s⁻²

To Find:

Thermal energy produced  = Eth = ?
Average friction force on the car = F = ?


Solution:

The thermal energy produced is due to work done (W) against the friction which is also equal to lose in the potential energy (P.E.) of the car. So we have the relation:

Eth = - W = - Loss in P.E. = - (P.Ef - P.Ei)

(-ve sign is due to work done against the friction)

Eth = - (mgh2 - mgh1)

Eth = - mgh2 + mgh1

Eth = mg (h1 - h2)

putting values

Eth = 1000 kg ｘ9.81 m s⁻² (40 m - 25 m)

Eth = 9810 kg m s⁻² (15 m)
 
Eth = 147,150 kg m² s⁻²

Eth = 147,150 J


Now to find the friction force

As we know that W = F . d so, 

F = `\frac {W}{d}`

by putting values

F = `\frac {147,150 J }{400 m}`

F = 367.875 N

or

F = 368 N -----------Ans

Thus the friction force is 368 N


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