Question

A small block of mass $200\text{g}$ is pressed against a horizontal spring fixed at one end to compress the spring through $5\text{cm}$. The spring constant is $100\text{N/m}$. The spring is located on a frictionless horizontal plank located at a height of $3\text{m}$ from the ground. When the block is released, what horizontal distance will it travel (after leaving the plank) before it hits the ground. (Take $g=10{\text{m/s}}^2$).

• A. $1\text{m}$
• B. $0.90\text{m}$
• C. $0.85\text{m}$
• D. $1.2\text{m}$

Answer 1

The correct option is C $0.85\text{m}$

Given :
$m=200\text{g}=0.2\text{kg},k=100\text{N/m},x=5\text{cm}=0.05\text{m},g=10{\text{m/s}}^2$

Taking horizontal plank as a reference.
Let the block hit the ground with velocity $v$.
From energy conservation :
$U_i+K_i=U_f+K_f$
$\frac{1}{2}k{x}^2=0+\frac{1}{2}m{v}^2$
$\Rightarrow v=\sqrt{\frac{100\times (0.05{)}^2}{0.2}}=1.11\text{m/sec}$

Applying equation of motion in $y$ direction
(Given : $3\text{m}$ below spring ) and take downward +ve
$3=ut+\frac{1}{2}g{t}^2$
$3=\frac{1}{2}g{t}^2\Rightarrow t=\sqrt{\frac{6}{g}}=0.77\text{s}$

$\therefore$ Displacement in x - direction
$d=u_x\times t+\frac{1}{2}{a}_x{t}^2$
$=u_{x}t$ (as $a_x=0$ in horizontal direction)
$d=u_x\times t=1.11\times 0.77$
$d=0.85\text{m}$