PE Integrated Concepts (a) Assuming 95.0% efficiency for the conversion of electrical power by the motor, what current must the 12.0-V batteries of a 750-kg electric car be able to supply: (a) To accelerate from rest to 25.0 m/s in 1.00 min? (b) To climb a 2.00 × 10? -m - high hill in 2.00 min at a constant 25.0-m/s speed while exerting 5.00 × 2? 10? N of force to overcome air resistance and friction? (c) To travel at a constant 25.0-m/s speed, exerting a 5.00×10? N force to overcome air resistance and friction? See Figure 20.44.

Step-by-step solution Given: Efficiency=95.0% Voltage = 12.0-V Mass = 750-kg Velocity = 25.0 m/s TIme= 1.00 min, 2.00 min Height = 2.00 × 10 -m Force = 5.00×10 N 2 1) Calculate the current supplied 12.0-V batteries of a 750-kg electric car be able to supply= 2) Calculate the current supplied To climb a 2.00 × 10 -m - high hill in 2.00 min at a constant 25.0-m/s speed while exerting 5.00×10 N of force to overcome air resistance and friction = 2 3) Calculate the current supplied to travel at a constant 25.0-m/s speed, exerting a 5.00×10 N force to overcome air resistance and friction= Step 1 of The power is defined as the rate flow of energy, P= E/t………………………(1) Here, E is the energy flowing and t is time duration for which energy flows. Also in an electric circuit, the rate at which energy is produced through a current carrying element is, P= I V………………………..(2) Here, I is the current and V is the voltage supplied. Step 2 of 6 Calculate the current supplied 12.0-V batteries of a 750-kg electric car be able to supply. The efficiency of the engine is 95%. So, the rate change of energy here is 95% of the power produced by the engine. 95% of P = 1/t (1/2mv ) 2 (0.95) IV = (1/2t mv ) 2 I =1/0.95 V (1/2t mv ) ………………….(3) Here, m is the mass of the car, t is time for which car travels and v is the velocity of car. Substitute 750 kg for m, 25.0 m/s for v, 12.0 V for V and 1.00 min for t.in equation (3) ,we get. 2 I= (750 kg )(25.0m/s) /2(0.95)(1.00min)(60s/1min)(12.0V) 2 I= (750 kg )(25.0m/s) /2(0.95)(60s)(12.0V) I= 343 A Hence, the current supplied is 343 A. 2 Step 3 of 6 Calculating current supplied to climb a 2.00 × 10 -m - high hill in 2.00 min at a constant 25.0-m/s speed while exerting 5.00×10 N of force to overcome air resistance and friction. The car climbs the tree with constant speed. So, there will be no change in its kinetic energy, thus the energy E is given as by equation (4) E= mgh + Fd…………………..(4) Here, F is the force applied to overcome resistance and d is the displacement of the car. The displacement of the car is calculated as given equation (5) as below d= vt……………….(5) Here, v is the speed of the car and t is the time taken by car to reach final height. Substitute vt for d in formula for energy. ½ mv2 - ½ mv = Pt ………0……(6)