Question

A pile of bricks of mass M is being raised to the tenth floor of a building of height H = 4y above the ground by a crane that is on top of the building. During the first part of the lift, the crane lifts the bricks a vertical distance h1=3y in a time t1=4T. During the second part of the lift, the crane lifts the bricks a vertical distance h2=y in t2=T. Which of the following correctly relates the power P1 generated by the crane during the first part of the lift to the power P2 generated by the crane during the second part of the lift?
A. P2=4P1
B. P2=43P1
C. P2=P1
D. P2=34P1
E. P2=13P1

Answer 1

Answer:

The correct option is  B

Explanation:

From the question we are told that

   The mass of the pile is  M

   The  height is  H  =  4 y

    The vertical distance achieve during the first lift is  $h_1 = 3 y$

     The time taken is  $t_1 = 4T$

    The vertical distance achieve during the second lift is  $h_2 = y$

     The time taken is $t_2 = T$

Generally the velocity of the crane during the first lift is  

      $v _1 = \frac{h_1}{t_1 }$

=>    $v _1 = \frac{3 y}{4T }$    

Generally the velocity of the crane during the second  lift is    

      $v _2 = \frac{h_2}{t_2 }$

=>    $v _2 = \frac{ y}{T}$  

Generally the power generated by  the crane during the first lift is    

    $P_1 = F_1 * v_1$

Here $F_1$ is the weight of the brick which is mathematically represented as

      $F_1 = M * g$ , g is the acceleration due to gravity

 So

       $P_1 = Mg * \frac{3y}{4T}$

Generally the power generated by  the crane during the first lift is    

    $P_1 = F_2 * v_2$

Here $F_2$ is the weight of the brick which is mathematically represented as

      $F_2 = M * g$ , g is the acceleration due to gravity

 So

       $P_1 = Mg * \frac{y}{T}$

The ratio of the first power generated to the second power is  

       $\frac{P_1}{P_2} = \frac{Mg * \frac{3y}{4T} }{ Mg * \frac{y}{T} }$

=>    $\frac{P_1}{P_2} = \frac{3}{4}$

=>   $P_2 = \frac{4}{3} P_1$