A 500 g stone is thrown up with a velocity of $15 ms^{-1}$. Find its
(i) P.E. at its maximum height
(ii) K.E. when it hits the ground
(56.25 J, 56.25 J)
Difficulty: Medium
Solution:
Mass of stone = m = 500 g = $\frac{500}{1000}$ kg = 0.5 kg
Velocity = v = 15 $ms^{-1}$
- Potential energy = P.E. = ?
- Kinetic energy = K.E. = ?
- Loss of K.E. = Gain in P.E.
$\frac{1}{2}m v_\_{f}^{2}$ – $\frac{1}{2}m v\_{i}^{2}$ = mgh
As the velocity of the stone at maximum height become zero, therefore, $v_{f}$ = 0
$\frac{1}{2}\times 0.5\times (0) -\frac{1}{2}\times 0.5 \times (15)^{2}$ = mgh
$\frac{-1}{2}\times 0.5 \times 225$ = mgh
$-$56.25 = mgh
mgh = $-$56.25J
Since energy is always positive, therefore
P.E. = 56.25 J
K.E. = $\frac{1}{2}m v ^{2}$
K.E. = $\frac{1}{2}\times 0.5 \times (15)^{2}$
= $\frac{1}{2}\times 0.5 \times 225$
= 56.25 J
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