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

Complete 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?

$A.\ \ P_2=4P_1$

$B.\ \  P_2=\frac{4}{3} P1$

$C.\ \  P_2=P_1$

$D. \ \ P_2=\frac{3}{4} P_1$

$E. \ \ \ P_2=\frac{1}{3} P_1$

Answer:

The correct option is  B

Explanation:

From the question we are told that  

    The mass of the brick is  M

    The  height height of the 10th floor is  H =  4y

     The height attained during the first part of the lift is  $h_1 =  3y$

     The time taken is $t_1 =  4T$

    The height attained during the second part of the lift is  $h_2  = y$

    The time taken is  $t_2  =  T$

 

Generally the velocity of the crane during the first lift is mathematically represented as

         $v_1  =  \frac{h_1}{t_1}$

=>      $v_1  =  \frac{3y}{4T}$

Generally the velocity of the crane during the first lift is mathematically represented as

         $v_1  =  \frac{h_2}{t_2}$

=>      $v_1  =  \frac{y}{T}$

Generally the power generated during the first lift is  

     $P_1 =  F_1 *  v_1$

Here $F_1$ force applied during the first lift which is mathematically represented as

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

So

       $P_1 =  Mg * \frac{3y}{4T}$

Generally the power generated during the second lift is  

     $P_2 =  F_2 *  v_2$

Here $F_2$ force applied during the second lift which is mathematically represented as

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

So

       $P_2 =  Mg * \frac{y}{T}$

So the ratio  of the first power to the second power is  

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

=>   $\frac{P_1}{P_2}  = \frac{3}{4}$

=>    $P_2 = \frac{4}{3} P_1$