First slide

Work Energy Theorem

Question

A small block of mass 0⋅1 kg is pressed against a horizontal spring fixed at one end to compress the spring through 5⋅0 cm as shown in figure.
The spring constant is 100 N/m. When released the block moves horizontally till it leaves the spring. Where will it hit the ground 2 m below the spring ?

Moderate

A

at a horizontal distance of 1 m from free end of the spring
B

at a horizontal distance of 2 m from free end of the spring
C

vertically below the edge on which the mass is resting
D

at a horizontal distance of $\sqrt{2} m$ from free end of the spring

Solution

The energy stored in spring will be given to mass as kinetic energy. Hence
$\begin{array}{l} \frac{1}{2} {kx}^{2} = \frac{1}{2} {mv}^{2} \\ \frac{1}{2} \times 100 \times {(\frac{5}{100})}^{2} = \frac{1}{2} \times 0 \cdot 1 \times v^{2} \end{array}$
Solving, we get $V = \sqrt{(5 / 2)}$
Now $y = 2 = 0 + \frac{1}{2} {gt}^{2}$
or $t = \sqrt{(4 / g)} = \sqrt{(2 / 5)}$
Further, $X = Vt = \sqrt{\frac{5}{2}} \times \sqrt{\frac{2}{5}} = 1 m$

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