AP EAMCET201923 Apr 2019Morning ShiftPhysicsWork, Power and EnergyActual
A constant power of (7 ~W ) is supplied on a toy car of mass (15 ~kg ). The distance travelled by the car when its velocity increases from (3 ~ms ⁻¹ ) to (5 ~ms ⁻¹ ) is
Options
- A(56 ~m )
- B(7 ~m )
- C(61 ~m )
- D(70 ~m )
Correct answer
D. (70 ~m )
Step-by-step solution
Given, (P=7 ~W ), mass, (m=15 ~kg ), (v_i=3 ~ms ⁻¹ and v_f=5 ~ms ⁻¹ ) From work-energy theorem, ( array rlrl & Work done & = 1 2 m (v_f^2-v_i^2 ) & & & = 1 2 15(25-9) & & W & =120 ~J array ) So, the time of work done, (t= work power = 120 7 =1714 ~s ) Hence, the acceleration, (a= v_f-v_i t = 5-3 17.14 =0.116 ~m / s ^2 ) Distance travelled by car, (D= v_f^2-v_i^2 2 a = 5^2-3^2 2 0.116 =68.96 70 ~m ) Hence, the correct option is (d).