Question

Calculate the Potential energy and distance covered by a stone of 5kg which is thrown upwards at a speed of 10m/s?

Answer 1

Solution :

⏭ Given:

✏ Mass of stone = 5kg

✏ Initial velocity of ball = 10mps (upward)

⏭ To Find:

• Potential energy at highest point
• Maximum height attained by ball

⏭ Concept:

✏ This question is completely based on concept of 'Energy Conversation'.

⏭ Calculation:

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• PE at highest point

$\mapsto \tt \: \red{U_i + K_i = U_f + K_f} \\ \\ \mapsto \sf \: 0 + \frac{1}{2} m {v}^{2} = U_f + 0 \\ \\ \mapsto \sf \: uf = \frac{1}{2} \times 5 \times {10}^{2} \\ \\ \mapsto \sf \: U_f = \dfrac{500}{2} \\ \\ \mapsto \boxed{ \tt{ \purple{U_f = 250 \: J}}}$

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• Max. height

$\leadsto \tt\: \green{U_f = m \times g \times H} \\ \\ \leadsto \sf \: 250 = 5 \times 10 \times H \\ \\ \leadsto \sf \: H = \dfrac{250}{50} \\ \\ \leadsto \: \boxed{ \tt{ \orange{Max. \: Height = 5 \: m}}}$

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⏭ Additional information:

• Energy is scalar quantity.
• Potential energy is relative quantity, means it requires reference point to define.

Answer 2

$\tt{\huge{\red{Hello!!! }}}$

$\tt{\green{Given \: - }}$

• Mass of stone = 5kg

• U_{ ball } = + 10m/s

We have to find the potential energy at the highest point and the maximum Height attained by the ball.

$PE _ { H P }\\ \\ \begin{lgathered}\mapsto \tt \: \blue{U_i + K_i = U_f + K_f} \\ \\ \mapsto \sf \: 0 + \frac{1}{2} m {v}^{2} = U_f + 0 \\ \\ \mapsto \sf \: uf = \frac{1}{2} \times 5 \times {10}^{2} \\ \\ \mapsto \sf \: U_f = \dfrac{500}{2} \\ \\ \mapsto \boxed{ \tt{ \orange{U_f = 250 \: J}}}\end{lgathered}$

$M_{ H } \\ \\ </p><p>\begin{lgathered}\leadsto \tt\: \purple{U_f = m \times g \times H} \\ \\ \implies \sf \: 250 = 5 \times 10 \times H \\ \\ \longrightarrow \sf \: H = \dfrac{250}{50} \\ \\ \disc \: \boxed{ \tt{ \blue{Max. \: Height = 5 \: m}}}\end{lgathered}$