Question

At a given time the momentum of a body of mass 5 kg is 10 kgms-1. Now a force or 0.2 N acts on the
body in the direction of motion for 10 seconds. The increase in kinetic energy is
(1) 2.2 J
(2) 3.3 J
(3) 4.4 J
(4) 5.5 J​

Answer 1

${ \huge \bf { \mid{ \overline{ \underline{Question}}} \mid}}$

At a given time the momentum of a body of mass 5 kg is 10 kgms-1. Now a force or 0.2 N acts on the body in the direction of motion for 10 seconds. The increase in kinetic energy is?

$\huge{\bold{\underline{\underline{ Answer}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Mass of the Body (m) = 5 kg.
• Initial Momentum (p) = 10 kg m/s.
• Force acting (F) = 0.2 N
• Time period (t) = 10 seconds.

$\huge{\bold{\underline{Explanation:-}}}$

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From Momentum Formula,

$\large{\boxed{\tt P = mu}}$

Substituting the values,

$\large{\tt \leadsto 10 \: Kg.m/s = 5 \: Kg \times u}$

$\large{\tt \leadsto 10 = 5 \times u}$

$\large{\tt \leadsto u = \dfrac{10}{5}}$

$\large{\tt \leadsto u = \cancel{\dfrac{10}{5}}}$

$\large{\leadsto {\underline{\underline{\tt u = 2 \: m/s}}}}$

Therefore, Initial velocity is 2 m/s.

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From Newton's second law of motion,

$\large{\boxed{\tt F = ma}}$

$\large{\tt \leadsto 0.2 \: N = 5 \: Kg \times a}$

$\large{\tt \leadsto 0.2 = 5 \times a}$

$\large{\tt \leadsto a = \dfrac{0.12}{5}}$

$\large{\tt \leadsto a = \cancel{\dfrac{0.2}{5}}}$

$\large{\leadsto {\underline{\underline{\tt a = 0.04 \: m/s^2}}}}$

Therefore, the Acceleration is 0.04 m/s².

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Applying First kinematics equation,

$\large{\boxed{\tt v = u + at}}$

Substituting the values,

$\large{\tt \leadsto v = 2 + 0.04 \times 10}$

$\large{\tt \leadsto v = 2 + 0.4}$

$\large{\leadsto {\underline{\underline{\tt v = 2.4 \: m/s}}}}$

Therefore, the Final velocity is 2.4 m/s.

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Applying Kinetic energy Formula,

$\large{\boxed{\tt \Delta K.E = \dfrac{1}{2}.m (v^2 - u^2)}}$

Substituting the values,

$\large{\tt \leadsto \Delta K.E = \dfrac{1}{2} \times 5 \times [(2.4)^2 - (2)^2]}$

$\large{\tt \leadsto \Delta K.E = \dfrac{1}{2} \times 5 \times [5.76 - 4]}$

$\large{\tt \leadsto \Delta K.E = \dfrac{1}{2} \times 5 \times 1.76}$

$\large{\tt \leadsto \Delta K.E = \dfrac{1}{\cancel{2}} \times 5 \times \cancel{1.76}}$

$\large{\tt \leadsto \Delta K.E = 1 \times 5 \times 0.88}$

$\large{\tt \leadsto \Delta K.E = 5 \times 0.88}$

$\huge{\boxed{\boxed{\tt \Delta K.E = 4.4 \: J}}}$

So, the Increase in Kinetic energy is 4.4 Joules (Option- 3).

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Answer 2

Answer:

Explanation:

(3) 3.3 J