Question 93: 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: