A block of mass $'m'$ is released from rest at point $A$. The compression in spring, when the speed of block is maximum
$\frac{{mg\sin \theta }}{k}$
$\frac{{2mg\sin \theta }}{k}$
$\frac{{mg\cos \theta }}{k}$
$\frac{{mg}}{k}$
An engine is attached to a wagon through a shock absorber of length $1.5\,m$. The system with a total mass of $50,000 \,kg$ is moving with a speed of $36\, km\,h^{-1}$ when the brakes are applied to bring it to rest. In the process of the system being brought to rest, the spring of the shock absorber gets compressed by $1.0\,m$.
If $90\%$ of energy of the wagon is lost due to friction, calculate the spring constant.
Two springs have their force constant as ${k_1}$ and ${k_2}({k_1} > {k_2})$. When they are stretched by the same force
The figure shows a mass $m$ on a frictionless surface. It is connected to rigid wall by the mean of a massless spring of its constant $k$. Initially, the spring is at its natural position. If a force of constant magnitude starts acting on the block towards right, then the speed of the block when the deformation in spring is $x,$ will be
Two blocks $A(5kg)$ and $B(2kg)$ attached to the ends of a spring constant $1120N/m$ are placed on a smooth horizontal plane with the spring undeformed. Simultaneously velocities of $3m/s$ and $10m/s$ along the line of the spring in the same direction are imparted to $A$ and $B$ then
$A$ block of mass $m$ moving with a velocity $v_0$ on a smooth horizontal surface strikes and compresses a spring of stiffness $k$ till mass comes to rest as shown in the figure. This phenomenon is observed by two observers:
$A$: standing on the horizontal surface
$B$: standing on the block
According to observer $B$, the potential energy of the spring increases