Question

A block of mass m is connected to a spring ( spring constant k ). Initially the block is at rest and the spring is in its natural length. Now the system is released in gravitational field and a variable force F is applied on the upper end of the spring such that the downward acceleration of the block is given as $a=g-\alpha t,$, where t is time elapses and $\alpha =1m/s^3,$ the velocity of the point of application of the force is

• A. $\frac{m}{k}-gt+\frac{t^2}{2}$
• B. $\frac{m}{k}+gt+\frac{t^2}{2}$
• C. $\frac{m}{k}-gt-\frac{t^2}{2}$
• D. $\frac{m}{k}-gt-t^2$

Answer 1

The correct option is A $\frac{m}{k}-gt+\frac{t^2}{2}$

Let the spring to be stretched to x at a time t, such that
$x=x_b+x_f$ where $x_{b}and{x}_f$ are elongation due to the body, m and Force, F.
Now, considering the body ,
mg-kx=ma
Or, $mg-kx=m(g-\alpha t)$
$kx=m\alpha t$
differentiating the above equation, we get
$\frac{d(x_b+x_f)}{dt}=\frac{m\alpha }{k}$
Or, $v_b+v_f=\frac{m\alpha }{k}\left(i\right)$
where $v_b$ and $v_f$ are the velocity of body m and the point of application of force.
Now, we have
$a=g-\alpha t$
Or, $\frac{d{v}_b}{dt}=g-\alpha t$
Integrating above, we get
$\int d{v}_b=\int (g-\alpha t)dt$
$v_b=gt-\frac{\alpha t^2}{2}\left(ii\right)$
From i and ii,
$v_f=\frac{m\alpha }{k}-gt+\frac{\alpha t^2}{2}$
($\because \alpha =1$) ,
$v_f=\frac{m}{k}-gt+\frac{t^2}{2}$