Question

A block with a mass of 6.0 kg is

Answer 1

(a) The normal force on the block is 50.92 N.

(b) The horizontal force on block keeping it in equilibrium is 29.4 N.

The given parameters;

• mass of the block, m = 6 kg
• angle of inclination, θ = 30.0°

The normal force on the block is calculated as follows;

$F_n = W \times cos(\theta)$

where;

• W is the weight of the block

$F_n = mg \times cos(\theta)\\\\F_n = 6 \times 9.8 \times cos(30)\\\\F_n = 50 .92 \ N$

The horizontal force on block keeping it in equilibrium is calculated as follows;

$F- F_x = 0\\\\F-mgsin\theta= 0\\\\F = mgsin\theta\\\\F = 6 \times 9.8 \times sin(30)\\\\F = 29.4 \ N$

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