High School

A boy pulls a 47.5 kg crate with a rope. The rope makes an angle of 28.0° to the horizontal.

The coefficient of kinetic friction for the crate and the deck is 0.300. The boy exerts a force of 185 N.

What is the acceleration of the crate?

Answer :

Final answer:

To calculate the acceleration of the crate, we divide the resultant force (the horizontal force exerted by the boy minus the frictional force) by the mass of the crate using Newton's Second Law.

Explanation:

The subjects of this problem are Newton's Second Law, the force of friction, and acceleration. To find the acceleration of the crate, we need to calculate the horizontal and vertical forces acting on it. The horizontal component of the force Fx is F.cosθ = 185N.cos(28°) and the vertical component Fy is F.sinθ = 185N.sin(28°). The weight of the crate W is m.g = 47.5kg*9.8m/s^2. The normal force N is W - Fy. The frictional force f due to kinetic friction is μk.N, where μk is the coefficient of kinetic friction, which in this case is 0.3. From here we use Newton's Second Law F = ma to find the acceleration: a = (Fx - f) / m.

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