How Much Force is Needed to Climb a Hill at Constant Speed with Air Resistance?

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To climb a 6-degree hill at a constant speed of 6 km/h, the bicyclist must overcome both gravitational force and air resistance. The gravitational force component along the hill is calculated as mg sin(6), resulting in 66.58 N. When accounting for air resistance, the total force required to maintain the same speed is 133 N. The model answer of 130 N may be due to rounding to two significant figures. No mistakes were made in the calculations presented.
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Homework Statement


A bicyclist of mass 65kg (including the bicycle) can coast down a 6 degree hill at steady speed of 6km/h because of air resistance. How much force must be applied to climb the hill at the same speed and same air resistance?


The Attempt at a Solution



since it is at constant speed. the net force is zero.

therefore the

Fg (force due to mg) - Fr (air resistance force) = 0

(mg sin 6) - Fr = 0

Fr= mg sin 6 = 66.58N.

For it to climb back the hill at the same speed and same air restance?

the bicyclist have to tackle the two same force now = 66.58 + mg sin 6 = 133N.

The model answer given is 130N...but i got 133N...

any mistake i have made?
 
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hemetite said:

Homework Statement


A bicyclist of mass 65kg (including the bicycle) can coast down a 6 degree hill at steady speed of 6km/h because of air resistance. How much force must be applied to climb the hill at the same speed and same air resistance?


The Attempt at a Solution



since it is at constant speed. the net force is zero.

therefore the

Fg (force due to mg) - Fr (air resistance force) = 0

(mg sin 6) - Fr = 0

Fr= mg sin 6 = 66.58N.

For it to climb back the hill at the same speed and same air restance?

the bicyclist have to tackle the two same force now = 66.58 + mg sin 6 = 133N.

The model answer given is 130N...but i got 133N...

any mistake i have made?
You haven't made any mistakes, perhaps the model answer has rounded to 2sf?
 
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