A $2$ kg body is dropped from height of $1$ m onto a spring of spring constant $800$ N/m as shown in the figure. A frictional force equivalent to $0.4$ kgwt acts on the body. The speed of the body just before striking the spring will be (Take $g = 10 m / s^{2}$ )

1 m/s

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2 m/s

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3 m/s

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4 m/s

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Solution

The correct option is B 4 m/s
We know that the change of kinetic energy is equal to work done by the system.

Change of Kinetic energy $= \frac{1}{2} m v^{2} - 0 = \frac{1}{2} m v^{2}$

Work done $= (m g - f) h$ , where $f = 0.4 k g w t = 0. 4 \times 10 = 4 N$ , friction force.

Now, $\frac{1}{2} m v^{2} = (m g - f) h$

$\frac{1}{2} 2 v^{2} = (20 - 4) 1$

$v^{2} = 16$
$v = 4 m / s$

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A 2 kg body is dropped from height of 1 m onto a spring of spring constant 800 N/m as shown in the figure. A frictional force equivalent to 0.4 kgwt acts on the body. The speed of the body just before striking the spring will be (Take g=10 m/s2)

The correct option is B 4 m/sWe know that the change of kinetic energy is equal to work done by the system. Change of Kinetic energy =12mv2−0=12mv2Work done =(mg−f)h , where f=0.4kgwt=0.4×10=4N, friction force.Now, 12mv2=(mg−f)h122v2=(20−4)1v2=16v=4 m/s

A $2$ kg body is dropped from height of $1$ m onto a spring of spring constant $800$ N/m as shown in the figure. A frictional force equivalent to $0.4$ kgwt acts on the body. The speed of the body just before striking the spring will be (Take $g = 10 m / s^{2}$ )