First slide

Work Energy Theorem

Question

A spring is compressed between two toy carts of masses $m_{1}$ and $m_{2}$ . When the toy carts are released the spring exerts on each toy cart equal and opposite forces for the same time t. If the coefficients of friction $μ$ between the ground and the toy carts are equal, then the displacements of the toy carts are in the ratio :

Moderate

A

$\frac{s_{1}}{s_{2}} = \frac{m_{2}}{m_{1}}$
B

$\frac{s_{1}}{s_{2}} = - \frac{m_{1}}{m_{2}}$
C

$\frac{s_{1}}{s_{2}} = - {(\frac{m_{2}}{m_{1}})}^{2}$
D

$\frac{s_{1}}{s_{2}} = - {(\frac{m_{1}}{m_{2}})}^{2}$

Solution

Let the minimum stopping distance be s. The force of friction would be p m g. Hence work done against friction
$W = μmgs Initial K.E. of toy cart = (p^{2} / 2 m) ∴ μmgs = (p^{2} / 2 m) or s = [p^{2} / (2 {μgm}^{2})] so s \propto 1 / m^{2} [.' p and μ ar e same for two toy carts . Further, the momentum is numerically same) \frac{s_{1}}{s_{2}} = {(\frac{m_{2}}{m_{1}})}^{2} As s_{1} and s_{2} are in opposite directions, so (\frac{s_{1}}{s_{2}}) = - {(\frac{m_{2}}{m_{1}})}^{2}$

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