Question:
A block of mass 50 kg is pushed on a frictionless inclined plane by a horizontal force F. The incline makes an angle of \(60^\circ\) with the horizontal. If the block moves with constant speed, then F is: (Take \(g=10\, m\,s^{-2}\))
A block of mass 50 kg is pushed on a frictionless inclined plane by a horizontal force F. The incline makes an angle of \(60^\circ\) with the horizontal. If the block moves with constant speed, then F is: (Take \(g=10\, m\,s^{-2}\))
Show Hint
For incline problems with horizontal force, always resolve forces along the plane using the angle between force direction and incline.
Updated On: Jun 19, 2026
- \(\frac{1000}{\sqrt{3}}\,N\)
- \(866\,N\)
- \(654\,N\)
- \(500\sqrt{2}\,N\)
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Step 1: Understand forces acting on the block.
The block is on a frictionless incline of \(60^\circ\). Forces acting are: - Weight \(mg\) vertically downward - Normal reaction \(N\) perpendicular to plane - Horizontal force \(F\) acting horizontally
Step 2: Resolve gravitational force along the incline.
Component of weight down the plane: \[ mg \sin 60^\circ = 50 \times 10 \times \frac{\sqrt{3}}{2} = 250\sqrt{3} \]
Step 3: Resolve horizontal force along incline.
Angle between horizontal force and incline is \(60^\circ\). So component of \(F\) along plane is: \[ F \cos 60^\circ = \frac{F}{2} \]
Step 4: Apply condition of constant speed.
Constant speed implies zero net force along incline: \[ F \cos 60^\circ = mg \sin 60^\circ \] \[ \frac{F}{2} = 250\sqrt{3} \]
Step 5: Solve for F.
\[ F = 500\sqrt{3} \approx 866\,N \]
Step 6: Final conclusion.
Thus, the required force is approximately: \[ 866\,N \]
Final Answer: \[ \boxed{866\,N} \]
The block is on a frictionless incline of \(60^\circ\). Forces acting are: - Weight \(mg\) vertically downward - Normal reaction \(N\) perpendicular to plane - Horizontal force \(F\) acting horizontally
Step 2: Resolve gravitational force along the incline.
Component of weight down the plane: \[ mg \sin 60^\circ = 50 \times 10 \times \frac{\sqrt{3}}{2} = 250\sqrt{3} \]
Step 3: Resolve horizontal force along incline.
Angle between horizontal force and incline is \(60^\circ\). So component of \(F\) along plane is: \[ F \cos 60^\circ = \frac{F}{2} \]
Step 4: Apply condition of constant speed.
Constant speed implies zero net force along incline: \[ F \cos 60^\circ = mg \sin 60^\circ \] \[ \frac{F}{2} = 250\sqrt{3} \]
Step 5: Solve for F.
\[ F = 500\sqrt{3} \approx 866\,N \]
Step 6: Final conclusion.
Thus, the required force is approximately: \[ 866\,N \]
Final Answer: \[ \boxed{866\,N} \]
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 Inclined planes Questions
- Sand is to be piled up on a horizontal ground in the form of a regular cone of a fixed base of radius R. Coefficient of static friction between the sand layers is p. Maximum volume of the sand that can be piled up in the form of cone without slipping on the ground is
- A \(3\,\text{kg}\) block on a rough incline of angle \(37^\circ\) is connected to a hanging mass of \(4\,\text{kg}\). If the coefficient of friction between the \(3\,\text{kg}\) block and the rough incline is \(\mu=0.25\), then the acceleration of the system is \[ g=10\,\text{m s}^{-2}, \qquad \sin37^\circ=0.6, \qquad \cos37^\circ=0.8 \]
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