Laws of Motion Quiz - ICSE Class 11 Physics
Free practice quiz: Laws of Motion (ICSE Class 11 Physics). 48 multiple-choice questions with answers and explanations. Play online, no sign-up needed.
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A block of mass $m$ is at rest on a rough inclined plane. If the angle of inclination is $\theta$, and the coefficient of static friction between the block and the plane is $\mu_s$, what is the condition for the block to remain at rest?
- $mg \sin \theta > \mu_s mg \cos \theta$
- $mg \sin \theta < \mu_s mg \cos \theta$
- $mg \sin \theta \le \mu_s mg \cos \theta$
- $mg \sin \theta = \mu_s mg \cos \theta$
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Answer: C. $mg \sin \theta \le \mu_s mg \cos \theta$
For the block to remain at rest, the component of gravity down the incline ($mg \sin \theta$) must be less than or equal to the maximum static friction force ($\mu_s N$, where $N = mg \cos \theta$ is the normal force). Thus, $mg \sin \theta \le \mu_s mg \cos \theta$.
A particle is moving in a straight line with constant acceleration. If its initial velocity is $u$ and its displacement after time $t$ is $s$, which of the following equations is NOT one of the standard kinematic equations?
- $s = ut + \frac{1}{2}at^2$
- $v = u + at$
- $v^2 = u^2 + 2as$
- $s = vt - \frac{1}{2}at^2$
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Answer: D. $s = vt - \frac{1}{2}at^2$
The standard kinematic equations for constant acceleration are $v = u + at$, $s = ut + \frac{1}{2}at^2$, and $v^2 = u^2 + 2as$. The equation $s = vt - \frac{1}{2}at^2$ is derived from $v = u+at \implies u = v-at$, and substituting this into $s = ut + \frac{1}{2}at^2$, giving $s = (v-at)t + \frac{1}{2}at^2 = vt - at^2 + \frac{1}{2}at^2 = vt - \frac{1}{2}at^2$. While it can be derived, it's typically not listed as a primary kinematic equation derived directly from definitions.
A block of mass $m$ is placed on a smooth horizontal surface. If a horizontal force $F$ is applied, the block accelerates. What is the acceleration of the block if a second identical block is placed next to it, and the same force $F$ is applied to the combined mass?
- Same as before
- Twice the previous acceleration
- Half the previous acceleration
- Depends on the distance between the blocks
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Answer: C. Half the previous acceleration
In the first case, $a_1 = F/m$. In the second case, the total mass is $2m$, and the force is still $F$. So, the acceleration is $a_2 = F/(2m) = \frac{1}{2} (F/m) = \frac{1}{2} a_1$. The acceleration is halved.
If a body moves with constant velocity, what is the work done by the net force on the body?
- Positive
- Negative
- Zero
- Cannot be determined
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Answer: C. Zero
Constant velocity implies zero acceleration. By Newton's second law ($F_{net} = ma$), if acceleration is zero, the net force must be zero. Work done by the net force is $W_{net} = F_{net} \cdot d$. If $F_{net} = 0$, then $W_{net} = 0$. Alternatively, the work-energy theorem states that the net work done equals the change in kinetic energy. If velocity is constant, kinetic energy is constant, so the change in kinetic energy is zero, hence the net work done is zero.
A block of mass $m$ is on a horizontal surface. A force $F$ is applied horizontally. If the block is just about to move, the frictional force is:
- Zero
- Equal to $F$
- Equal to $\mu_s N$
- Greater than $F$
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Answer: B. Equal to $F$
When a body is at rest, the static friction force adjusts itself to be equal in magnitude and opposite in direction to the applied force, up to a maximum value ($\mu_s N$). If the block is 'just about to move', it means the applied force $F$ is equal to the maximum static friction, and the static friction force itself is equal in magnitude to this applied force.