An electric lift with a maximum load of 2000 kg (lift+ passengers) is moving up with a constant speed of 1.5 ms -1. The frictional force opposing the motion is 3000 N. The minimum power delivered by the motor to the lift in watts is: (g=10 ms -2)

Question

An electric lift with a maximum load of 2000 kg (lift+ passengers) is moving up with a constant speed of 1.5 ms -1. The frictional force opposing the motion is 3000 N. The minimum power delivered by the motor to the lift in watts is: (g=10 ms -2) (A) 23500 (B) 23000 (C) 20000 (D) 34500

Tardigrade Admin · Accepted Answer

Constant velocity ⇒ a =0 ⇒ T = W + f =20000+3000 =23000 N <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/08c9e5dbc38d36bbeab13891c8e3143d-.png" /> ⇒ Power = Tv =23000 × 1.5 =34500 watts

Q. An electric lift with a maximum load of $2000 k g$ (lift+ passengers) is moving up with a constant speed of $1.5 m s^{- 1}$ . The frictional force opposing the motion is $3000 N$ . The minimum power delivered by the motor to the lift in watts is : $(g = 10 m s^{- 2})$

23500

23000

20000

34500

Constant velocity $\Rightarrow a = 0$
$\Rightarrow T = W + f$
$= 20000 + 3000$
$= 23000 N$

$\Rightarrow$ Power $= T v$
$= 23000 \times 1.5$
$= 34500$ watts

Q. An electric lift with a maximum load of 2000kg (lift+ passengers) is moving up with a constant speed of 1.5ms−1. The frictional force opposing the motion is 3000N. The minimum power delivered by the motor to the lift in watts is : (g=10ms−2)