Question

A system of wedge and block as shown in figure, is released with the spring in its natural length. All surfaces are frictionless. Maximum elongation in the
spring will be
(A)2mg sin theta/K
(B)mgsin theta /K
(C)4mgsin theta/K
(D)mgsin theta/K
​

Answer 1

Formulae to be used:-

• Work done a constant force = $\sf (\vec{F}.\vec{s}) = F\times s\times cos(\theta)$
• Work done by a spring = $\sf W_S = -\dfrac{1}{2}K(x_f^2 - x_i^2),$ where K is spring constant, $\sf x_i$ is the initial compression/elongation of the spring, and $\sf x_f$ is the final compression/elongation of the spring.

Answer:-

Let the maximum elongation of the spring be $\sf x.$

Refer to the attachment

In $\sf \triangle ABC,$

$\sf sin(\theta) = \dfrac{h}{x}$

$\longrightarrow \sf h = x\;sin(\theta)\quad\quad\dots (1)$

Now, we will use work energy theorem

According to work energy theorem,

Work done by all forces on an object will be equal to the change in the kinetic energy.

$\sf W_{(All\:forces)} = \Delta K.E. = K.E._f - K.E._i$

• In this problem, as the block is released from rest, $\sf K.E._i = 0\:J$
• And, maximum elongation of spring will happen only when the block stops somewhere along the wedge (we have assumed that to be $\sf x$.) So, $\sf K.E._f = 0\:J$
• The only two forces acting on the block are - it's own weight and the spring force.
• As the spring was in natural position, $\sf x_i = 0$
• As the maximum elongation is asked, which we have assumed to be $\sf x,$ $\sf x_f = x$

So,

$\longrightarrow \sf W_{(mg)} + W_S = 0 - 0$

$\longrightarrow \sf W_{(mg)} = - W_S$

$\longrightarrow \sf mg\times h \times cos(0^o) = - \bigg\{ - \dfrac{1}{2}K(x_f^2 - x_i^2) \bigg\}$

$\longrightarrow \sf mgh = \dfrac{1}{2}K(x^2 - 0^2)$

$\longrightarrow \sf mgh = \dfrac{1}{2}Kx^2$

From (1),

$\longrightarrow \sf mgx\:sin(\theta) = \dfrac{1}{2}Kx^2$

$\longrightarrow \sf mg\:sin(\theta) = \dfrac{1}{2}Kx$

$\longrightarrow \sf \underline{\underline{\sf{\green{ x = \dfrac{2mg\:sin(\theta)}{K} }}}}$