System Parameters:

• Mass of block, $m = 1.0 \text{ kg}$
• Initial velocity of block, $\vec{v}_0 = -10 \hat{i} \text{ m/s}$ (Moving left)
• Velocity of belt, $\vec{v}_{belt} = +10 \hat{i} \text{ m/s}$ (Moving right)
• Coefficient of kinetic friction, $\mu_k = 0.5$

1. Force and Acceleration

The relative velocity of the block with respect to the belt is:

$$ \vec{v}_{rel} = \vec{v}_{block} – \vec{v}_{belt} = -10 – 10 = -20 \text{ m/s} $$

Since relative motion is to the left, kinetic friction acts to the right.

$$ f_k = \mu_k mg = 0.5 \times 1.0 \times 10 = 5 \text{ N} $$

Acceleration of the block:

$$ \vec{a} = \frac{f_k}{m} \hat{i} = \frac{5}{1} \hat{i} = +5 \text{ m/s}^2 $$

2. Time to Stop Slipping

Slipping stops when the block’s velocity matches the belt’s velocity ($10 \text{ m/s}$).

$$ v = u + at \implies 10 = -10 + 5t $$ $$ 20 = 5t \implies t = 4 \text{ s} $$

3. Evaluating the Matches

(a) Work done by friction on the block with respect to the ground ($W_{f,g}$)

Displacement of block w.r.t ground in $t=4$ s:

$$ s_g = ut + \frac{1}{2}at^2 = -10(4) + \frac{1}{2}(5)(4)^2 = -40 + 40 = 0 \text{ m} $$

Work done:

$$ W_{f,g} = \vec{f} \cdot \vec{s}_g = 5 \times 0 = 0 \text{ J} $$

Match: (a) $\rightarrow$ (p)

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(b) Change in momentum of the block with respect to the ground ($\Delta P$)

$$ \Delta P = m(v_f – v_i) = 1.0 \times (10 – (-10)) = 20 \text{ kg m/s} $$

Match: (b) $\rightarrow$ (t) (Note: (t) is visible in the full context as 20)

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(c) Distance slid by the block on the belt

This is the magnitude of the relative displacement.

$$ s_{rel} = u_{rel}t + \frac{1}{2}a_{rel}t^2 $$

Where $u_{rel} = -20 \text{ m/s}$ and $a_{rel} = a_{block} – a_{belt} = 5 – 0 = 5 \text{ m/s}^2$.

$$ s_{rel} = -20(4) + \frac{1}{2}(5)(16) = -80 + 40 = -40 \text{ m} $$

Distance slid = $|-40| = 40 \text{ m}$.

Match: (c) $\rightarrow$ (s)

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(d) Work done by friction on the block with respect to the belt

In the belt frame, force is $+5 \text{ N}$ and displacement is $-40 \text{ m}$.

$$ W_{rel} = \vec{f} \cdot \vec{s}_{rel} = 5 \times (-40) = -200 \text{ J} $$

Modulus is $200$.

Additionally, let’s check work done by friction on the belt w.r.t ground. Force on belt is $-5 \text{ N}$. Displacement of belt is $10 \times 4 = 40 \text{ m}$. Work = $-200 \text{ J}$. Modulus is $200$.

Match: (d) $\rightarrow$ (q) and (r) (Assuming q/r correspond to 200 based on options).