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A small block of mass m is kept on a wedge of M, which is kept on ground,there is no friction anywhere.aswer the following :
1. Acceleration of M w.r.t. ground will be a = mgsintheta costheta/xM+msin^2theta then the value of x is
2. The value of normal force applied on M will be N = xMmg costheta/ M+msin^2theta then the value of x is ?
1. Acceleration of M w.r.t. ground will be a = mgsintheta costheta/xM+msin^2theta then the value of x is
2. The value of normal force applied on M will be N = xMmg costheta/ M+msin^2theta then the value of x is ?
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A small block of mass m is kept on a wedge of M, which is kept on grou...
1. Acceleration of M w.r.t. ground
The acceleration of the block of mass m w.r.t. the ground can be calculated using Newton's second law. The forces acting on the block of mass m are its weight mg acting downwards and the normal reaction N acting perpendicular to the incline. The block is accelerating along the incline with an acceleration 'a'. The forces acting on the wedge of mass M are its weight Mg acting downwards and the normal reaction acting perpendicular to the ground. The forces acting on the wedge are horizontal acceleration due to block m and the normal force at the surface of contact. The net force acting on the wedge is given by:
$Mg - N = Ma$
The net force acting on the block is given by:
$mgsin\theta - N = ma$
Solving these two equations simultaneously, we get:
$a = \frac{mgsin\theta}{M + msin^2\theta}$
2. Normal force applied on M
The normal force acting on the wedge can be calculated by substituting the value of acceleration 'a' in the equation for the net force acting on the wedge:
$Mg - N = \frac{mgsin\theta}{M + msin^2\theta}$
Solving for N, we get:
$N = Mg - \frac{mgsin\theta}{M + msin^2\theta}$
Which can be further simplified to:
$N = \frac{xMmgcos\theta}{M + msin^2\theta}$
Thus, the value of x in both cases is equal to M.
The acceleration of the block of mass m w.r.t. the ground can be calculated using Newton's second law. The forces acting on the block of mass m are its weight mg acting downwards and the normal reaction N acting perpendicular to the incline. The block is accelerating along the incline with an acceleration 'a'. The forces acting on the wedge of mass M are its weight Mg acting downwards and the normal reaction acting perpendicular to the ground. The forces acting on the wedge are horizontal acceleration due to block m and the normal force at the surface of contact. The net force acting on the wedge is given by:
$Mg - N = Ma$
The net force acting on the block is given by:
$mgsin\theta - N = ma$
Solving these two equations simultaneously, we get:
$a = \frac{mgsin\theta}{M + msin^2\theta}$
2. Normal force applied on M
The normal force acting on the wedge can be calculated by substituting the value of acceleration 'a' in the equation for the net force acting on the wedge:
$Mg - N = \frac{mgsin\theta}{M + msin^2\theta}$
Solving for N, we get:
$N = Mg - \frac{mgsin\theta}{M + msin^2\theta}$
Which can be further simplified to:
$N = \frac{xMmgcos\theta}{M + msin^2\theta}$
Thus, the value of x in both cases is equal to M.
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A small block of mass m is kept on a wedge of M, which is kept on ground,there is no friction anywhere.aswer the following :1. Acceleration of M w.r.t. ground will be a = mgsintheta costheta/xM+msin^2theta then the value of x is 2. The value of normal force applied on M will be N = xMmg costheta/ M+msin^2theta then the value of x is ?
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A small block of mass m is kept on a wedge of M, which is kept on ground,there is no friction anywhere.aswer the following :1. Acceleration of M w.r.t. ground will be a = mgsintheta costheta/xM+msin^2theta then the value of x is 2. The value of normal force applied on M will be N = xMmg costheta/ M+msin^2theta then the value of x is ? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A small block of mass m is kept on a wedge of M, which is kept on ground,there is no friction anywhere.aswer the following :1. Acceleration of M w.r.t. ground will be a = mgsintheta costheta/xM+msin^2theta then the value of x is 2. The value of normal force applied on M will be N = xMmg costheta/ M+msin^2theta then the value of x is ? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A small block of mass m is kept on a wedge of M, which is kept on ground,there is no friction anywhere.aswer the following :1. Acceleration of M w.r.t. ground will be a = mgsintheta costheta/xM+msin^2theta then the value of x is 2. The value of normal force applied on M will be N = xMmg costheta/ M+msin^2theta then the value of x is ?.
A small block of mass m is kept on a wedge of M, which is kept on ground,there is no friction anywhere.aswer the following :1. Acceleration of M w.r.t. ground will be a = mgsintheta costheta/xM+msin^2theta then the value of x is 2. The value of normal force applied on M will be N = xMmg costheta/ M+msin^2theta then the value of x is ? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A small block of mass m is kept on a wedge of M, which is kept on ground,there is no friction anywhere.aswer the following :1. Acceleration of M w.r.t. ground will be a = mgsintheta costheta/xM+msin^2theta then the value of x is 2. The value of normal force applied on M will be N = xMmg costheta/ M+msin^2theta then the value of x is ? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A small block of mass m is kept on a wedge of M, which is kept on ground,there is no friction anywhere.aswer the following :1. Acceleration of M w.r.t. ground will be a = mgsintheta costheta/xM+msin^2theta then the value of x is 2. The value of normal force applied on M will be N = xMmg costheta/ M+msin^2theta then the value of x is ?.
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