Question

At a certain point potential energy and kinetic energy of a body of mass 0.1kg moving vertically up are 3.6 joules and 6.2 joules. It rises to a maximum height of ________m
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Answer 1

Answer:

10 metres

Explanation:

Logic to be used: At the maximum height the overall energy gets converted into potential energy, i.e, while moving up the kinetic energy of the body (6.2 J) is lost and gets converted into P.E.

mgh= 6.2 + 3.6

1/10 x 9.8 x h = 9.8

Therefore h = 10 metres

Answer 2

Answer:

10 m

Explanation:

Concept

• Potential energy arises in systems with parts that exert forces on each other of a magnitude dependent on the configuration, or relative position, of the parts.
• The kinetic energy is the measure of the work that an object does by virtue of its motion. Simple activities like walking, jumping, throwing, and falling involves kinetic energy. In this article, let us familiarize ourselves with the concept of kinetic energy.

Given

P.E=3.6J

K.E=6.2J

g=9.8m/s²

m=0.1kg

Find

Maximum height

Solution

Potential energy,

$P.E =m g h=3.6 \mathrm{~J}$

$\mathrm{h}=\frac{3.6}{0.1 \times 9.8}=3.67 \mathrm{~m}$

Kinetic energy,

$\mathrm{K} . \mathrm{E}=\frac{1}{2} \mathrm{mv} \mathrm{v}^{2}=6.2 \mathrm{~J}$

$\mathrm{v}^{2}=\frac{6.2 \times 2}{0.1}$

$v^{2}=124$

3rd equation of motion,

$\mathrm{v}^{2}-\mathrm{u}^{2}=-2 \mathrm{hg}$

$124-u^{2}=-9.8 \times 3.67 \times 2$

$u^{2}=194.56 (initial velocity)$

For maximum height, $\mathrm{v}=0 \mathrm{~m} / \mathrm{s}$

$v^{2}-u^{2}=-2 h_{m} g$

$h_{m}=\frac{194.56}{2 \times 9.8}$

$h_{\mathrm{m}}=10 \mathrm{~m}$

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