A 800 kg car starts from rest at a height of 15 m above the bottom of the hill.

Question

A 800 kg car starts from rest at a height of 15 m above the bottom of the hill. He accelerates down hill & reaches a speed of 20 m/s at the bottom of the hill (lowest point in the figure) where it runs out of gasoline. Then the car continues coasting up another hill until it comes to rest. Ignore air resistance & all frictional forces. Note that constant acceleration equations cannot be applied to this problem since the acceleration is not necessarily constant. a) The work done by the from start until the car runs of gas is: b) The highest position the car reaches above the bottom of the second hill is: d) The momentum of the car at the bottom of the hill is: d) Suppose instead of coasting upon the second hill, the driver applies the breaks & stops the car at a height of 8 m above the lowest point. The energy dissipated by the breaks in this case is: e) The latent heat of melting of ice is 80 cal/g. If none of the heat generated by the breaks is lost, the energy dissipates by the breaks would melt _____ g of ice which is already at 0 degree C

Explanation / Answer

PART-1: At the height of 15m, the car will have potential energy only,

P = m*g*h1 = 800kg*10m/s2*15m = 12*104J

At the bottom of the hill where car runs out of gas, it will have only kinetic energy,

K = (1/2)*m*v2= (1/2)*800kg*(20m/s)2 = 16*104 J

Work done by the engine at the bottom is the diffrence in two energies,

W = 16*104 - 12*104 = 4*104 J

PART-2: At the highest point that car reaches above the bottom of the second hill, it will have potential energy only, P = m*g*h2

Now applying law of conservation of energy at the bottom at the highest point reaches by the car on second hill,

K1 + P1 = K2 + P2

16*103 + 0 = 0 + m*g*h2 = 800kg*10m/s2*h2

h2 = 20 m (highest point the car reaches on the second hill after it runs out of the gasoline)

PART-3: Momentum at the bottom the hill, p = m*v = 800kg*20m/s = 16000 kgm/s

PART-4: Energy dissipated by the breaks = energy at the height of 20m - energy at the height of 8m

= m*g*h20m - m*g*h8m = m*g*(h20m - h8m)

= 800kg*10m/s2*( 20 - 8) = 96*103 J

PART-5: Energy dissipated by the breaks = 96*103 J = (96*103 J/4.186 J) cal = 22.9*103 cal

Now, heat absorbed by the ice, Q = m*Lice

  m = Q/Lice = (22.9*103 cal) / (80cal/gm) = 286.7gm

  

  

  

  

  

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