Question

A 1,300-N crate is being pushed across a level floor at a constant speed by a force F of 390 N at an angle of 20.0° below the horizontal, as shown in the figure a below. Two figures show a side view of a crate positioned upon a horizontal surface. Figure (a): An arrow pointing downward and to the right is labeled vector F and forms an angle of 20° below the horizontal as it approaches the upper left edge of the crate. Figure (b): An arrow pointing upward and to the right is labeled vector F and forms an angle of 20° above the horizontal as it extends from the upper right edge of the crate.

(a) What is the coefficient of kinetic friction between the crate and the floor? (Enter your answer to at least three decimal places.)

0.256 Correct: Your answer is correct.

(b) If the 390-N force is instead pulling the block at an angle of 20.0° above the horizontal, as shown in the figure b, what will be the acceleration of the crate? Assume that the coefficient of friction is the same as that found in part (a).

0.7 Incorrect: Your answer is incorrect.

Answer 1

1.

Applied force, F = 390 N

Weight of the crate, W = mg = 1300 N

Angle at which the force applied, θ = 20⁰ below the horizontal

Horizontal component of force F = F Cos θ = 390 x Cos 20 = 366.48 N

Vertically downward component of force F = F Sin θ = 390 x sin 20 = 133.38 N

Friction, f = μN = μ ( mg + F Sin θ)

To maintain a constant speed, f = F Cos θ

μ ( mg + F Sin θ) = F Cos θ

μ x ( 1300 + 133.38 ) = 366.48

μ = 366.48/( 1300 + 133.38 ) = 0.256

the coefficient of kinetic friction, μ = 0.256

2.

Applied force, F = 390 N

Weight of the crate, W = mg = 1300 N

Angle at which the force applied, θ = 20⁰ above the horizontal

Horizontal component of force F = F Cos θ = 390 x Cos 20 = 366.48 N

Vertically upward component of force F = F Sin θ = 390 x sin 20 = 133.38 N

Friction, f = μN = μ ( mg - F Sin θ) = 0.256 x ( 1300 – 133.38) = 298.65 N

Net force, F_net = F Cos θ – f = 366.48 – 298.65 = 67.83

Mass of the crate, m = W/g = 1300/9.8 = 132.35 kg

If a is the acceleration of the crate then,

, F_net = ma

a = F_net /m = 67.83/132.35 = 0.512 m/s²

acceleration of the crate, a = 0.512 m/s²