Question 15: Sliding Block on Irregular Path
Step 1: The Physics Principle
Since the block starts from rest and comes to a halt, the net work done on the block is zero according to the Work-Energy Theorem.
$$ W_{gravity} + W_{friction} = \Delta K = 0 $$
Step 2: Calculating Work
Let the starting point be $A(0, -2)$ and the stopping point be $P(x, y)$.
- Work by Gravity: $W_g = mg(y_i – y_f) = mg(-2 – y)$
- Work by Friction: For a path with negligible curvature ($a_c \approx 0$), the normal force is $N = mg \cos \theta$. The horizontal component of displacement is $dx = ds \cos \theta$. $$ W_f = \int -\mu N ds = \int -\mu (mg \cos \theta) \frac{dx}{\cos \theta} = -\mu mg x $$
Step 3: The Equation of Motion
Equating the sum of work to zero:
$$ mg(-2 – y) – \mu mg x = 0 $$
$$ -2 – y = 0.6 x $$
$$ y = -0.6x – 2 $$
We must find the point $(x,y)$ on the provided graph that satisfies this linear equation.
Step 4: Verification
Testing Option (b) $x=10$:
$$ y = -0.6(10) – 2 = -8 $$
The point $(10, -8)$ lies perfectly on the intersection of our calculated friction-work line and the graphical trajectory.