Question 70: Laws of Motion

Question

A block of mass $$m$$ is placed on a surface having vertical crossection given by $$y=x^2 / 4$$. If coefficient of friction is 0.5, the maximum height above the ground at which block can be placed without slipping is:

Accepted Answer

Given:
The equation of the surface: $$y = \frac{x^2}{4}$$.
Coefficient of friction: $$\mu = 0.5$$.
Gravitational acceleration: $$g$$ (assumed constant).
1.
Slope of the Surface
:
The slope of the surface at any point is given by:
$$ 	an	heta = \frac{dy}{dx} $$
From the equation of the surface:
$$ y = \frac{x^2}{4} $$
Differentiating with respect to $x$:
$$ \frac{dy}{dx} = \frac{x}{2} $$
Thus, the slope at any point is:
$$ 	an	heta = \frac{x}{2} $$
2.
Forces Acting on the Block
:
At the point where the block is placed:
Weight
of the block acts vertically downward: $ mg $.
Normal force
acts perpendicular to the surface.
Frictional force
acts parallel to the surface, opposing the component of the weight that causes slipping.
3.
Condition for No Slipping
:
The block will not slip if the frictional force is sufficient to counteract the component of the gravitational force parallel to the slope. The frictional force is:
$$ f = \mu N $$
The normal force $N$ is given by:
$$ N = mg \cos	heta $$
The component of weight parallel to the slope is:
$$ F_{	ext{parallel}} = mg \sin	heta $$
For the block to not slip:
$$ f \geq F_{	ext{parallel}} $$
Substitute $f = \mu N$ and $N = mg \cos	heta$:
$$ \mu (mg \cos	heta) \geq mg \sin	heta $$
Simplify:
$$ \mu \cos	heta \geq \sin	heta $$
Divide through by $\cos	heta$:
$$ \mu \geq 	an	heta $$
4.
Maximum Slope Without Slipping
:
From the above condition:
$$ 	an	heta \leq \mu $$
Substitute $\mu = 0.5$:
$$ 	an	heta \leq 0.5 $$
Thus:
$$ \frac{x}{2} \leq 0.5 $$
$$ x \leq 1 $$
5.
Maximum Height
:
The height $y$ of the block at $x = 1$ is obtained from the surface equation:
$$ y = \frac{x^2}{4} $$
Substitute $x = 1$:
$$ y = \frac{1^2}{4} = \frac{1}{4} \, 	ext{m} $$
Final Answer:
The maximum height above the ground at which the block can be placed without slipping is:
$$ \boxed{\frac{1}{4} \, 	ext{m}} $$
Thus, the correct option is
Option D
.

Answer

1/2 m

Answer

1/3 m

Answer

1/6 m

Answer

1/4 m

Laws of Motion

A block of mass $m is placed on a surface having vertical crossection given by y=x^2 / 4$. If coefficient of friction is 0.5, the maximum height above the ground at which block can be placed without slipping is:

Choose the correct answer: