Solution to Question 20
A block is sliding on a horizontal floor.
- Friction ($f_k$): Always acts opposite to the direction of velocity. Magnitude $f_k = \mu mg$.
- Applied Force ($F$): The problem states “a constant horizontal force is applied on it opposite to its direction of motion.” This implies a vector $\vec{F}$ that initially points opposite to $\vec{v}$.
Phase 1: Initial Motion (slowing down)
The block is moving to the right ($v > 0$).
Forces acting:
1. Friction (Left): $f_k$
2. Applied Force (Left): $F$
Net Force: $F_{net} = F + f_k$ (acting Left).
Acceleration: $a_1 = -\frac{F + f_k}{m}$.
The magnitude of deceleration is high. The velocity-time graph will have a steep negative slope.
Phase 2: At Moment of Stop ($v = 0$)
The block comes to rest. The applied force $F$ is constant and continues to act to the Left.
If $F > (f_s)_{max}$ (static friction limit), the block will start moving to the Left.
Phase 3: Reversed Motion
The block now moves to the Left ($v < 0$).
Forces acting:
1. Applied Force (Left): $F$ (driving the motion).
2. Friction (Right): $f_k$ (opposing motion).
Net Force: $F_{net} = F – f_k$ (acting Left).
Acceleration: $a_2 = \frac{F – f_k}{m}$ (to the Left).
Note that $|a_2| < |a_1|$ because friction now subtracts from the applied force instead of adding to it.
The velocity-time graph must show a line that crosses the time axis (reverses direction). The slope before crossing (deceleration) must be steeper than the slope after crossing (acceleration in reverse).
Graph (a) perfectly matches this description. If $F < (f_s)_{max}$ then the block stops at equilibrium this matches option (d)