Question

Consider the situation shown in figure. The elevator is going up with an acceleration of 2.00 m s−2 and the focal length of the mirror is 12.0 cm. All the surfaces are smooth and the pulley is light. The mass-pulley system is released from rest (with respect to the elevator) at t = 0 when the distance of B from the mirror is 42.0 cm. Find the distance between the image of the block B and the mirror at t = 0.200 s. Take g = 10 m s−2.
Figure

Answer 1

Explanation:

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Answer 2

The distance between the image of the block B and the mirror is 8.57 cm.

Explanation:

Let a = acceleration of masses A and B from the free body diagram (w.r.t elevator)

T- mg + ma - 2m = 0   -----> eqn (1)

Consequently,

T - ma = 0   -----> eqn (2)

From equation 1 and 2 ,

− + − 2 =   −

− + −2 = −

2 − 2 − = 0

2 = 2 +

2 = 2 +

$\begin{equation}a=\frac{m(2+g)}{2 m}\end$

g is given

$\begin{equation}g=10 \mathrm{ms}^{-2}\end$

$\begin{equation}a=\frac{(2+10)}{2}\end$

$\begin{equation}a=\frac{12}{2}\end$

a = 6 m/s²

Thus, the distance traveled in t = 0.2 s by B,  

$\begin{equation}S=\frac{1}{2} a t^{2}\end$

$\begin{equation}S=\frac{1}{2} \times 6 \times 0.2^{2}\end$

S = 3 × 0.04

S = 0.12 m

S = 12 cm  

So far from mirror

u = -(42-12) = -30 cm

f = 12 cm  

From Equation of mirrors ,

$\begin{equation}\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\end$

$\begin{equation}\frac{1}{v}+\frac{1}{-30}=\frac{1}{12}\end$

$\begin{equation}\frac{1}{v}=\frac{1}{12}+\frac{1}{30}\end$

$\begin{equation}\frac{1}{v}=\frac{5+2}{60}\end$

$\begin{equation}\frac{1}{v}=\frac{7}{60}\end$

$\begin{equation}v=\frac{60}{7}\end$

v = 8.57 cm  

Thus, the distance between block B and mirror image = 8.57 cm.