a 25 kg crate rests on the level floor. a horisontal force of 50 n accelerate the crate at 1m/s2 calculate the normal force of the crate, the frictional force on the crate, and the coefficient of kinetic frietion between the crate and floor

Question

contestada

Respuesta :

PiaDeveau PiaDeveau · Answer 1 · 2020-06-16T04:14:24+02:00

Answer:

Normal force on the crate = 245 N

Coefficient of kinetic (μ) friction between the crate and the floor = 0.102

Frictional force on the crate (Ff) = 25 N

Explanation:

Given:

Mass of crate (m) = 25 kg

Horizontal force (F) = 50 N

Acceleration (a) = 1 m/s²

Find:

1. Normal force on the crate .

2. Coefficient of kinetic (μ) friction between the crate and the floor.

3. Frictional force on the crate (Ff) .

Computation:

1. Normal force on the crate .

Normal force on the crate = Weight

Weight(w) = mg [ ∵ g = 9.8 ]

Weight = (25)(9.8)

Weight = 245 N

Normal force on the crate = 245 N

2. Coefficient of kinetic (μ) friction between the crate and the floor.

(Normal force)μ = ma

245(μ) = (25)(1)

245(μ) = 25

μ = 0.102

Coefficient of kinetic (μ) friction between the crate and the floor = 0.102

3. Frictional force on the crate (Ff).

Frictional force on the crate (Ff) = (Normal force)μ

Frictional force on the crate (Ff) = 245(0.102)

Frictional force on the crate (Ff) = 25 N