High School

A block of mass 20 kg rests on a rough plane inclined at an angle of θ to the horizontal, such that [tex]\frac{7}{25} \sin \theta = 1[/tex]. The coefficient of friction between the block and the plane is 0.2.

(i) Find the force, acting parallel to the plane, required to stop the block from sliding down the plane.

(ii) Find the force, acting parallel to the plane, required to move the block up the plane.

Options:
A. (i) 196 N, (ii) 44 N
B. (i) 44 N, (ii) 196 N
C. (i) 196 N, (ii) 196 N
D. (i) 44 N, (ii) 44 N

Answer :

The problem requires determining the forces needed to keep a block from sliding down an inclined plane and to move it up, involving calculations with the coefficient of friction and the incline angle. So, none of the options are correct.

The student is dealing with a physics problem involving statics and dynamics on an inclined plane, often encountered in high school curricula. Specifically, the student seeks to determine the forces required to prevent a block from sliding down an inclined plane and move it up again, given the coefficient of friction and the incline angle.

To resolve this problem, one must first calculate the angle of the incline using the provided equation 7/25 sin(θ) = 1. Solving for sin(θ) gives sin(θ) = 25/7, which is not possible as the sine of an angle cannot exceed 1. Therefore, it seems there's an error in the provided equation. Assuming a typo and that the correct equation might be 7/25 = sin(θ), we find θ ≈ 16.26°.

The force required to prevent the block from sliding down the plane (force of static friction) can be found by calculating the component of gravitational force parallel to the incline and equating it to the maximum static friction force, i.e., F = μ_s * N, where N = mg cos(θ) and μ_s is the coefficient of static friction. For moving the block up the plane, the required force must overcome both the component of gravitational force down the incline and the force of kinetic friction, which is μ_k * N.