A battery-operated car utilizes a 120.0 V battery with negligible internal resistance. Find the charge, in coulombs, the batteries must be able to store and move in order to accelerate the 770 kg car from rest to 26 m/s, make it climb a 2.15 x 10^2 m high hill while maintaining that speed, and then cause it to travel at a constant 26 m/s by exerting a 5.3 x 10^2 N force for an hour

4.29×10⁵ C

Explanation:

From the question,

The energy stored in the battery = Kinetic energy of the car+ Energy needed to make the car climbed the hill+Energy required to exert a force.

E = 1/2mv²+mgh+Fd Equation 1

Where E = Energy stored in the battery, m = mass of the car, v = velocity of the car, h = height of the hill, F = force exerted on the car, d = distance traveled by the car.

But,

d = vt Equation 2

Where v = velocity, t = time.

Substitute equation 2 into equation 1

E = 1/2mv²+mgh+F(vt) Equation 3

Given: m = 770 kg, v = 26 m/s, h = 2.15×10² m = 215 m, F = 5.3×10² N = 530 N, t = 1 hour = 3600 s, g = 9.8 m/s²

Substitute into equation 1

E = 1/2(770)(26²)+(770)(9.8)(215)+(530)(26)(3600)

E = 260260+1622390+49608000

E = 51490650 J

Using,

E = qV Equation 4

Where q = charge of the battery, V = Voltage.

make q the subject of the equation

q = E/V Equation 5

Given: E = 51490650 J, V = 120 V

Substitute into equation 5

q = 51490650/120

q = 429088.75 C

q = 4.29×10⁵ C

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