The magnitude of the work done by the force of friction is 1045.68 joules
Explanation:
Work done = force × distance
A 16.7 kg block is dragged over a rough, horizontal surface by a constant force of 132 N acting at an angle of 25.8. above the horizontal. The block is displaced 57,8 m, and the coefficient of kinetic friction is 0.229
To find the normal reaction force of the block distribute The constant force into two component, horizontal component of 132 cos(25.8°) and vertical component 132 sin(25.8°), Then equate ∑vertical forces by 0 to find R
The weight of the block = mg, where m is its mass and g is 9.8 m/s²
→ ∑Vertical forces = R + 132 sin(25.8°) - mg
→ ∑Vertical forces = R + 132 sin(25.8°) - (16.7)(9.8)
→ ∑Vertical forces = R - 106.2
Equate it by 0
→ R - 106.2 = 0 ⇒ add 106.2 to both sides
→ R = 106.2 N
Now let us find the friction force
→ F = μ R
→ μ = 0.229
→ F = 0.229 (106.2) = 24.3198 N
Let us calculate the work don by the force of friction
→ W = F × d
→ d = 57.8 meters
→ W = 24.3198 × 57.8 = 1045.68 joules
The magnitude of the work done by the force of friction is 1045.68 joules
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