Q.36. A spherical ball of mass falls from a height of and is brought to rest after penetrating in

Q: What is the correct answer for this exam Physics Q36?

The correct answer is B. Using the work-energy theorem: $W_{\text{total}} = \Delta K$. Work done by gravity: $W_g = mg(h_1 + h_2)$, where $h_1 = 10 \mathrm{~m}$ (fall height) and $h_2 = 10 \mathrm{~cm} = 0.1 \mathrm{~m}$ (penetration depth). $W_g = 2 \times 10 \times (10 + 0.1) = 2 \times 10 \times 10.1 = 202 \mathrm{~J}$. Work done by sand: $W_{\text{sand}} = -F_{\text{sand}} \times h_2 = -F_{\text{sand}} \times 0.1$. Th

Q: Q.36. A spherical ball of mass falls from a height of and is brought to rest after penetrating into sand. The average force exerted by sand on the ball is __________ .

The correct answer is B. Using the work-energy theorem: $W_{\text{total}} = \Delta K$. Work done by gravity: $W_g = mg(h_1 + h_2)$, where $h_1 = 10 \mathrm{~m}$ (fall height) and $h_2 = 10 \mathrm{~cm} = 0.1 \mathrm{~m}$ (penetration depth). $W_g = 2 \times 10 \times (10 + 0.1) = 2 \times 10 \times 10.1 = 202 \mathrm{~J}$. Work done by sand: $W_{\text{sand}} = -F_{\text{sand}} \times h_2 = -F_{\text{sand}} \times 0.1$. Th

Q: What are the options for this this exam Physics question?

A. 1980 · B. 2020 · C. 2000 · D. 1000

Question

Q.36. A spherical ball of mass  falls from a height of  and is brought to rest after penetrating  into sand. The average force exerted by sand on the ball is __________ .

FirstInTest · Accepted Answer

Using the work-energy theorem: $W_{	ext{total}} = \Delta K$.
Work done by gravity: $W_g = mg(h_1 + h_2)$, where $h_1 = 10 \mathrm{~m}$ (fall height) and $h_2 = 10 \mathrm{~cm} = 0.1 \mathrm{~m}$ (penetration depth).
$W_g = 2 	imes 10 	imes (10 + 0.1) = 2 	imes 10 	imes 10.1 = 202 \mathrm{~J}$.
Work done by sand: $W_{	ext{sand}} = -F_{	ext{sand}} 	imes h_2 = -F_{	ext{sand}} 	imes 0.1$.
The initial kinetic energy is $K_i = 0$ (starts from rest). The final kinetic energy is $K_f = 0$ (comes to rest).
So, $\Delta K = K_f - K_i = 0$.
$W_g + W_{	ext{sand}} = 0$
$202 - F_{	ext{sand}} 	imes 0.1 = 0$
$F_{	ext{sand}} = \frac{202}{0.1} = 2020 \mathrm{~N}$.

Answer

1980

Answer

2020

Answer

2000

Answer

1000

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