Question

In the situation shown a block of mass 2 kg is constrained to move perpendicular to the incline plane of the wedge. If the wedge is moving with acceleration 5 m/s2 horizontally, the normal force between block and wedge is (assume no friction)


{ Only serious answers please }​

Answer 1

Answer:

force is mass × acceleration

so mass ha 2

a ha 5

ta 5 duni10

Answer 2

Answer:

The Normal force during the acceleration of wedge between block and wedge plane is 7.6669 Newton

Explanation:

Provided that:

Mass of the block (m) = 2 Kg

Inclination of the wedge plane (θ) = 37°

Horizontal acceleration of the Wedge plane (a) = 5 m/sec^2

If the wedge would be stationary then Normal force between block and wedge:

(N1) = mg Cosθ N

$N1 = mg \cosθ \\ = 2 \times 9.8 \times \cos 37 \degree \: newton \\ = 15.653 \: newton$

When not acceleration the normal force is 15.653 Newton.

But as the wedge is moving with acceleration;

The Norma force between two surfaces will be decreased.

The vertical component of acceleration will decrease the Normal force;

A supporting figure is also given;

So now Normal force:

$N2= mg \cosθ - ma \sin(90 - \theta) \\ N2=mg \cos37 \degree - ma \sin 53 \degree \\ = (2 \times 9.8 \times cos37 \degree ) - (2 \times 5 \times \sin 53 \degree) \: newton\\ = 7.6669 \: newton$

So the Normal force during the acceleration of wedge between block and wedge plane is 7.6669 Newton.