Question:
A block of 10 kg mass is connected to the top of a frictionless inclined plane of inclination \(45^\circ\) by a spring of negligible mass, unstretched length 0.8 m, and spring constant \(100\sqrt{2} \, \text{N/m}\). Find the distance of the block from the top along the incline in the equilibrium position. (Acceleration due to gravity \(g = 10 \, \text{m/s}^2\))
A block of 10 kg mass is connected to the top of a frictionless inclined plane of inclination \(45^\circ\) by a spring of negligible mass, unstretched length 0.8 m, and spring constant \(100\sqrt{2} \, \text{N/m}\). Find the distance of the block from the top along the incline in the equilibrium position. (Acceleration due to gravity \(g = 10 \, \text{m/s}^2\))
Show Hint
For a block attached to a spring on an incline, at equilibrium use \(mg \sin \theta = k (x - l_0)\) to find displacement along the incline.
Updated On: Jul 18, 2026
- 0.9 m
- 1.1 m
- 1.3 m
- 1.5 m
Show Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Step 1: Resolve forces along the incline.
At equilibrium, the component of gravity along the incline is balanced by the spring force:
\[ mg \sin \theta = k (x - l_0) \]
where \(x\) is the distance along incline, \(l_0 = 0.8 \, \text{m}\) is natural length of the spring, \(k = 100\sqrt{2} \, \text{N/m}\).
Step 2: Substitute known values.
\[ 10 \cdot 10 \cdot \sin 45^\circ = 100\sqrt{2} (x - 0.8) \]
\[ 100 \cdot \frac{\sqrt{2}}{2} = 100\sqrt{2} (x - 0.8) \]
Step 3: Simplify equation.
\[ 50 \sqrt{2} = 100 \sqrt{2} (x - 0.8) \]
\[ x - 0.8 = \frac{50 \sqrt{2}}{100 \sqrt{2}} = 0.5 \]
Step 4: Solve for \(x\).
\[ x = 0.8 + 0.5 = 1.3 \, \text{m} \]
Step 5: Verify units and reasoning.
Spring force and component of gravity both in N, distance in meters, equilibrium condition satisfied.
Step 6: Final conclusion.
Hence, the block is located at:
\[ \boxed{1.3 \, \text{m}} \]
At equilibrium, the component of gravity along the incline is balanced by the spring force:
\[ mg \sin \theta = k (x - l_0) \]
where \(x\) is the distance along incline, \(l_0 = 0.8 \, \text{m}\) is natural length of the spring, \(k = 100\sqrt{2} \, \text{N/m}\).
Step 2: Substitute known values.
\[ 10 \cdot 10 \cdot \sin 45^\circ = 100\sqrt{2} (x - 0.8) \]
\[ 100 \cdot \frac{\sqrt{2}}{2} = 100\sqrt{2} (x - 0.8) \]
Step 3: Simplify equation.
\[ 50 \sqrt{2} = 100 \sqrt{2} (x - 0.8) \]
\[ x - 0.8 = \frac{50 \sqrt{2}}{100 \sqrt{2}} = 0.5 \]
Step 4: Solve for \(x\).
\[ x = 0.8 + 0.5 = 1.3 \, \text{m} \]
Step 5: Verify units and reasoning.
Spring force and component of gravity both in N, distance in meters, equilibrium condition satisfied.
Step 6: Final conclusion.
Hence, the block is located at:
\[ \boxed{1.3 \, \text{m}} \]
Was this answer helpful?
0
0
Top AP EAPCET Physics Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The radii of the two columns in a U - tube are r1 and r2. When a liquid of density p (angle of contact is $ 0{}^\circ $ ) is filled in it, the level difference of the liquid in the two arms is h. The surface tension of the liquid is: (g = acceleration due to gravity)
- A rigid metallic sphere is spinning around its own axis in the absence of external torque. If the temperature is raised, its volume increases by $9 \%$. The change in its angular speed is
View More Questions
Top AP EAPCET Simple Harmonic Motion Questions
- A body is executing simple harmonic motion. At a displacement \( x \), its potential energy is \( E_1 \), and at a displacement \( y \), its potential energy is \( E_2 \). The potential energy \( E \) at a displacement \( (x + y) \) is:
- A block A of mass 1.6 kg is executing SHM with amplitude of 10 cm. A block B of mass 0.9 kg is kept on A. When A passes through its equilibrium, the amplitude of oscillation of the system of block A and B is
- A particle is executing simple harmonic motion. If the minimum time taken by the particle to move from extreme position to half of the amplitude is \( t_1 \), and the minimum time taken by the particle to move from mean position to half of the amplitude is \( t_2 \), then
- When a body of mass 8 kg is attached to a spring balance, the reading of the balance is 20 cm. Instead of 8 kg, if another body of mass M is suspended from the spring balance and is made to oscillate vertically, the time period of oscillation is \( \frac{\pi}{5} \) s, then the value of M is (Acceleration due to gravity = 10 ms\(^{-2} \))
- Under the action of a force \( F = -75y \) where F is in Newton and y is in meters, an object of mass 3 kg executes simple harmonic motion. If the velocity of the object at the mean position is 2.5 ms\(^{-1} \), the maximum acceleration of the object is
View More Questions
Top AP EAPCET Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- The four quantum numbers of the valence electron of potassium are :
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
View More Questions