Q26E - University Physics with Modern Physics

Question

A mass m slides down a smooth inclined plane from an initial vertical height h, making an angle a with the horizontal. 

(a) The work done by a force is the sum of the work done by the components of the force. Consider the components of gravity parallel and perpendicular to the surface of the plane. Calculate the work done on the mass by each of the components, and use these results to show that the work done by gravity is exactly the same as if the mass had fallen straight down through the air from a height h. 

(b) Use the work–energy theorem to prove that the speed of the mass at the bottom of the incline is the same as if the mass had been dropped from height h, independent of the angle a of the incline. Explain how this speed can be independent of the slope angle. 

(c) Use the results of part (b) to find the speed of a rock that slides down an icy frictionless hill, starting from rest 15.0 m above the bottom.

Step-by-Step Solution

Answer

1. The work done due to gravitational force when the body falls from a height h is similar to work is done by gravity if a mass m slides down a smooth inclined plane from a starting vertical height h.
2. The velocity of the block to independent of the slope is $\sqrt{2gh}$.
3. The velocity of rock to slide down the hill is 17.15 m/s.

Step 1: Identification of given data

The given data can be listed below,

• The height of the inclined plane is, h 
• The angle of inclination is, $\alpha$

Step 2: Concept/Significance of gravitational force

Any two mass-containing objects are attracted to one another by the gravitational pull. Because it consistently attempts to bring masses together rather than pushing them apart, the gravitational force is known as attractive.

Step 3: Determination of the work done on the mass by each of the components, and use these results to show that the work done by gravity is exactly the same as if the mass had fallen straight down through the air from a height h

The sum of forces perpendicular to the inclined plane is given by,

 

 $$\begin{gathered} N+\left(-mg \cos \alpha \right)=0 \\ N=mg \cos \alpha \end{gathered}$$

 

The work done due to gravity is given by,

 

$$\begin{gathered} W_g=F_{g}s \\ =mgs \sin \alpha \end{gathered}$$

 

The free body diagram of the system is shown below as,

 

 

From the figure, it is clear that the height of the plane is given by,

 

$h=s \sin \alpha$

 

Substitute this value in the equation for work done as:

 

$W_g=mgh$ 

 

When the body directly falls from height h the work done due to gravitational force is given by,

 

 $$\begin{gathered} W_g={\int }_{y1}^{y2}{F}_{g}ds \\ ={\int }_{y1}^{y2}{F}_{g}ds \cos \varphi \end{gathered}$$

 

Here $F_g$ is the gravitational force, s is the inclined distance, $y_1$is the initial height, and $y_2$ is the final height of the inclined plane.

 

Substitute all the values in the above,

$$\begin{gathered} W_g={\int }_h^0{F}_{g}ds \cos 180° \\ ={\int }_h^0{F}_{g}ds \\ =F_{g}h \\ =mgh \end{gathered}$$ 

 

Thus, from the calculation, it is clear that the work done due to gravitational force when the body falls from a height h is similar to work is done by gravity if a mass m slides down a smooth inclined plane from a starting vertical height h.

Step 4: (b) Explanation of how the speed can be independent of the slope angle

The sole factor affecting how much work gravity does is an object's vertical displacement. There is a tiny force component in the direction of the displacement, yet a high displacement in this direction when the slope angle is minimal. The force component in the direction of the displacement along the incline is bigger, but the displacement in this direction is lower when the slope angle is considerable.

 

The total work done for the block is given by,

$W_{tot}=\frac{1}{2} mv\frac{2}{2}-\frac{1}{2} m{v}_1^2$ 

 

Here, m is the mass of the block, vis the final velocity of the block, $v_1$ is the initial velocity of the block.

 

Substitute all the values the velocity is given by,

$$\begin{gathered} mgh=\frac{1}{2}m{v}_2^2-0 \\ =\sqrt{2gh} \end{gathered}$$

 

Thus, the velocity of the block to independent of the slope is $\sqrt{2gh}$.

Step 5: (c) Determination of the speed of a rock that slides down an icy frictionless hill, starting from rest 15.0 m above the bottom

The velocity of the rock sliding down is given by,

 

$v=\sqrt{2gh}$

 

Here is the acceleration due to gravity, and h is the height of the slope.

 

Substitute all the values in the above,

 

$$\begin{gathered} v=\sqrt{2\left(9.8 m/s^2\right)\left(15 m\right)} \\ =\sqrt{294 m^2/s^2} \\ =17.15 m/s \end{gathered}$$

 

Thus, the velocity of rock to slide down the hill is 17.15 m/s.