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------------------------------------------------ A 17.8 kg block is dragged over a rough, horizontal surface by a constant force of 141 N acting at an angle of 28.3 degrees above the horizontal. The block is displaced 62.3 m, and the coefficient of kinetic friction is 0.136. Find the work done by the 141 N force.

a) [tex]7.8 \times 10^3 \, \text{J}[/tex]
b) [tex]1.6 \times 10^4 \, \text{J}[/tex]
c) [tex]3.2 \times 10^4 \, \text{J}[/tex]
d) [tex]6.4 \times 10^4 \, \text{J}[/tex]

The work done by the 141 N force acting at an angle of 28.3 degrees above the horizontal on a block over a displacement of 62.3 m is approximately 7.8×10³ joules.

Explanation:

The work done by the 141 N force on a 17.8 kg block can be calculated by using the formula W = F × d × cos(θ), where W is the work done, F is the force, d is the displacement, and θ is the angle of the force relative to the displacement. The horizontal component of the force does the work, which is F × cos(θ). Plugging in the values gives us W = 141 N × 62.3 m × cos(28.3°). The cosine of 28.3° is approximately 0.881. Therefore, the work done is 141 N × 62.3 m × 0.881 ≈ 7.8×10³ J, which corresponds to option a).