Question

5.A block with a mass 0.25 kg sitting on a frictionless surface is connected to a light spring with spring constant 180 N/m .If the block is displaced 15 cm from its equilibrium position and released.The Total Energy and its speed (respectively) of the block when it is 10 cm from equilibrium position is;. Single choice.
(5 Points)
2.0 J and 3.0 m/s
4.05 J and 9 m/s
1.0J and maximum speed
16J and 25 m/s

Answer 1

Solution

When block will displaced 0.15m then released

Then energy wiil be stored in spring so

$E=\frac{1}{2}kx^2\\E=\frac{1}{2}\times180\times0.15^2$

$E=2.0J$

At x=0.10m it will have some kinetic energy and some potential energy

So by energy conservation

$\frac{1}{2}k(x_1^2-x_2^2)=\frac{1}{2}mv^2$

So by putting the value

$180(0.15^2-0.1^2)=0.25v^2$

$v=\sqrt{9}=3m/s$

So option (a) is correct.