Question

A 3.2 kg block starts at rest and slides a distance d down a frictionless 30. $0^{\circ }$ incline , where it runs into a spring . The block slides an addition 21.0 cm before it is brought to rest momentarily by compressing the spring , whose spring constant is 431 $\frac{N}{m}$

what is the value of d ?

• A. 0.08m
• B. 0.18m
• C. 0.28m
• D. 0.38m

Answer 1

The correct option is D 0.38m

All the forces on block will be

(1) Gravity

(2) Spring force

(3) Normal

Let 'v' be the velocity with which the block collides with the spring
$W_g$ + $W_N$ + $W_{sp}$ = $\Delta$ KE = 0

Normal is always perpendicular to displacement so $W_N$ = 0
$W_{mg}+W_{spring}=0-\frac{1}{2}m{v}^2\left(mgsin\theta \right)\left(21cm\right)-\frac{1}{2}K(21cm{)}^2=-\frac{1}{2}m{v}^2\text{substituting the above values}\Rightarrow v=\sqrt{\frac{2\times 6.1435}{3.2}}\therefore \text{If we consider the motion of block through a distance 'd' units}Applying{v}^2-u^2=2as{v}^2-0^2=2(gsin\theta )d\Rightarrow d=\frac{v^2}{2gsin\theta }\text{substituting the respective values in the above equation}\therefore d=0.38m$