A block of mass m is placed on a smooth wedge of inclination θ. The whole system is accelerated horizontally so that the block does not slip on the wedge. The force exerted by the wedge on the block (g is acceleration due to gravity) will be[2004]a)mg/cosθb)mg cosθc)mg sinθd)mgCorrect answer is option 'A'. Can you explain this answer?

Question

A block of mass m is placed on a smooth wedge of inclination &theta;. The whole system is accelerated horizontally so that the block does not slip on the wedge. The force exerted by the wedge on the block (g is acceleration due to gravity) will be[2004]a)mg/cos&theta;b)mg cos&theta;c)mg sin&theta;d)mgCorrect answer is option 'A'. Can you explain this answer?

Answer

Answer

Answer

Θ, as shown in the figure below. The wedge is given an acceleration a to the right. Find the acceleration of the block and the normal force acting on it.

![image.png](attachment:image.png)

To solve this problem, we need to resolve the forces acting on the block and the wedge. We can assume that the system is frictionless, so there is no force of friction acting on the block.

Let F be the force acting on the wedge in the horizontal direction. Then, the force acting on the block in the horizontal direction is Fcosθ, and the force acting on the block in the vertical direction is Fsinθ.

The weight of the block is mg, where g is the acceleration due to gravity. The normal force acting on the block is N.

We can write the equations of motion for the block and the wedge as follows:

For the block:

ma = Fcosθ - Nsinθ

For the wedge:

Ma = F - Ncosθ

where M is the mass of the wedge.

We can eliminate N from the above equations by multiplying the first equation by cosθ and adding it to the second equation:

Ma + macosθ = F

Substituting a = a'cosθ, where a' is the acceleration of the block in the horizontal direction, we get:

Ma' + Mgsinθ = F

Solving for a', we get:

a' = (F - Mgsinθ)/M

Substituting F = Ma + Nsinθ and N = Mgsinθ - Ma/tanθ in the above equation, we get:

a' = gsinθ/(1 + M/m*sin^2θ)

The normal force acting on the block can be found by substituting a' in the equation for the block's motion:

N = mgcosθ - ma'

Substituting the value of a', we get:

N = mgcosθ - (F - Mgsinθ)/M

Substituting F = Ma + Nsinθ and solving for N, we get:

N = (M + m)gcosθ/(1 + M/m*sin^2θ)

Therefore, the acceleration of the block is a' = gsinθ/(1 + M/m*sin^2θ), and the normal force acting on it is N = (M + m)gcosθ/(1 + M/m*sin^2θ).

Answer

Mg/cos theta option A

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