Physics, 19.10.2020 17:01 erinmcdonald6673

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

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Answer from: jacobdobson5856

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?

$A.\ \ P_2=4P_1$

$B.\ \ P_2=\frac{4}{3} P1$

$C.\ \ P_2=P_1$

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

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

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

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

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

$P_2 = Mg * \frac{y}{T}$

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

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

=> $\frac{P_1}{P_2} = \frac{3}{4}$

=> $P_2 = \frac{4}{3} P_1$

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