A ball of mass 300 g is dropped from a height | vedclass

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(B) The mass of the ball is $m = 300 g = 0.3 kg$. The total height fallen by the ball is $H = 10 m + 1.5 m = 11.5 m$. According to the work-energy the

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$23$

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(B) The mass of the ball is $m = 300 g = 0.3 kg$. The total height fallen by the ball is $H = 10 m + 1.5 m = 11.5 m$. According to the work-energy theorem,the total work done by all forces is equal to the change in kinetic energy. Since the ball starts from rest and comes to rest,the change in kinetic energy is $\Delta K = 0$. The forces acting on the ball are gravity (downward) and the resistance force of the sand (upward). Work done by gravity $W_g = mgH = 0.3 	imes 10 	imes 11.5 = 34.5 J$. Work done by sand resistance $W_R = -F_R 	imes d$,where $d = 1.5 m$. Applying the theorem: $W_g + W_R = 0 \Rightarrow 34.5 - F_R 	imes 1.5 = 0 \Rightarrow F_R = \frac{34.5}{1.5} = 23 N$.

$A$ ball of mass $300 g$ is dropped from a height $10 m$ above a sandy ground. On reaching the ground,it penetrates through a distance $1.5 m$ in sand and finally stops. The average resistance offered by the sand to oppose the motion is (acceleration due to gravity $= 10 m s^{-2}$) (in $N$)

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