- #1

simphys

- 324

- 46

- Homework Statement
- 96. Proper design of automobile braking systems must account

for heat buildup under heavy braking. Calculate the thermal energy dissipated from brakes in a 1500-kg car that descends a 17° hill. The car begins braking when its speed is ##95km/h## (##=26.39m/s##)and slows to a speed of ##35km/h##(##=9.72m/s##) in a distance of 0.30 km measured along the road.

- Relevant Equations
- conservation of enertgy

My sign doesn't check out and I don't get why that'd be the case.

Forces that act --> ##F_{fr} and F_g##

derivation:

##\Delta K = W_{NC} + W_C (1)##

##\Delta K + \Delta U = W_{NC}##

##\frac 12 mv_2^2 - \frac 12 mv_1^2 + mgy_2 - mgy_1 = W_{NC}## NOTE: ##y_2## assumed to be datum line so ##y_2 == 0##

filling in the data I get:

##W_{NC} = -1.742E6 = E_{th}## instead of ##1.742E6 ##

Do I need to make ##W_{NC}## to ##-W_{NC}## in ##(1)## perhaps or?

or..

can I just say ##W_{NC} = -1.742E6## which is negative work done by the friction force.

From this we are able to conclude that the

Forces that act --> ##F_{fr} and F_g##

derivation:

##\Delta K = W_{NC} + W_C (1)##

##\Delta K + \Delta U = W_{NC}##

##\frac 12 mv_2^2 - \frac 12 mv_1^2 + mgy_2 - mgy_1 = W_{NC}## NOTE: ##y_2## assumed to be datum line so ##y_2 == 0##

filling in the data I get:

##W_{NC} = -1.742E6 = E_{th}## instead of ##1.742E6 ##

Do I need to make ##W_{NC}## to ##-W_{NC}## in ##(1)## perhaps or?

or..

can I just say ##W_{NC} = -1.742E6## which is negative work done by the friction force.

From this we are able to conclude that the

**magnitude**of the Thermal energy dissipated ##E_th = 1.742E6##. Is this permitted or do I actually need to get a + sign in the equation itself?
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