Work Energy Theorem and Conservation of Mechanical Energy Questions in English

(B) According to the work-energy theorem,the net work done on the ball is equal to the change in its kinetic energy.
Since the ball starts from rest and ends at rest,the change in kinetic energy is $0$.
The forces acting on the ball are gravity $(mg)$ and the resistive force of the sand $(F_{avg})$.
The total displacement of the ball is $H + d$,where $H = 10 \text{ m}$ and $d = 10 \text{ cm} = 0.1 \text{ m}$.
Work done by gravity + Work done by sand = Change in kinetic energy
$mg(H + d) - F_{avg} \times d = 0$
$F_{avg} = \frac{mg(H + d)}{d}$
Substituting the values: $F_{avg} = \frac{2 \times 10 \times (10 + 0.1)}{0.1} = \frac{20 \times 10.1}{0.1} = 200 \times 10.1 = 2020 \text{ N}$.

(C) The work done by the force is equal to the change in kinetic energy: $W = \Delta K = \frac{1}{2}m(v_f^2 - v_i^2)$.
Given $v = 2x^2$,at $x = 0$,the initial velocity $v_i = 2(0)^2 = 0 \text{ m/s}$.
At $x = 5 \text{ m}$,the final velocity $v_f = 2(5)^2 = 2(25) = 50 \text{ m/s}$.
Substituting the values into the work-energy theorem:
$W = \frac{1}{2} \times 1 \text{ kg} \times ((50 \text{ m/s})^2 - (0 \text{ m/s})^2)$
$W = \frac{1}{2} \times 1 \times 2500 = 1250 \text{ J}$.

(A) Using the principle of conservation of mechanical energy,the loss in potential energy of the $6 \text{ kg}$ mass is equal to the gain in kinetic energy of the system (both $6 \text{ kg}$ and $2 \text{ kg}$ masses) plus the gain in potential energy of the $2 \text{ kg}$ mass.
Let $m_1 = 6 \text{ kg}$ and $m_2 = 2 \text{ kg}$. When the $6 \text{ kg}$ mass falls by $h = 6 \text{ m}$,the $2 \text{ kg}$ mass rises by $h = 6 \text{ m}$.
Loss in potential energy of $m_1 = m_1 g h = 6 \times 10 \times 6 = 360 \text{ J}$.
Gain in potential energy of $m_2 = m_2 g h = 2 \times 10 \times 6 = 120 \text{ J}$.
Gain in kinetic energy of the system = $\frac{1}{2}(m_1 + m_2)v^2 = \frac{1}{2}(6 + 2)v^2 = 4v^2$.
By conservation of energy: $m_1 g h = m_2 g h + \frac{1}{2}(m_1 + m_2)v^2$.
$360 = 120 + 4v^2$.
$240 = 4v^2$.
$v^2 = 60$.
$v = \sqrt{60} \approx 7.746 \text{ m/s}$.

(D) According to the work-energy theorem,the total work done on the object is equal to the change in its kinetic energy: $W_{\text{total}} = \Delta K$.
Here,the total work done is the sum of the work done by gravity $(W_g)$ and the work done by the resistive force $(W_r)$: $W_g + W_r = \Delta K$.
Given: mass $m = 1 \text{ g} = 0.001 \text{ kg}$,height $h = 1 \text{ km} = 1000 \text{ m}$,initial velocity $u = 0 \text{ m/s}$,final velocity $v = 5 \text{ m/s}$,and $g = 10 \text{ m/s}^2$.
Work done by gravity: $W_g = mgh = 0.001 \text{ kg} \times 10 \text{ m/s}^2 \times 1000 \text{ m} = 10 \text{ J}$.
Change in kinetic energy: $\Delta K = \frac{1}{2}mv^2 - \frac{1}{2}mu^2 = \frac{1}{2} \times 0.001 \text{ kg} \times (5 \text{ m/s})^2 - 0 = 0.0005 \times 25 = 0.0125 \text{ J}$.
Substituting the values into the work-energy theorem: $10 \text{ J} + W_r = 0.0125 \text{ J}$.
$W_r = 0.0125 \text{ J} - 10 \text{ J} = -9.9875 \text{ J}$.
The closest option is $-9.98 \text{ J}$.