To move a large crate across a rough floor, you push on it with a force F at an angle of 21°, below the horizontal, as shown in the figure. Find the acceleration of the crate, given that the mass of the crate is m = 33 kg, the applied force is 322 N and the coefficient of kinetic friction between the crate and the floor is 0.62.

m/s2

a = 0.865 m / s²

Explanation:

For this exercise we use Newton's second law

X axis

          Fₓ - fr = ma

Y Axis

         N- W- = 0

We use trigonometry for the components of the force

        sin 21 = F_{y} / F

        F_{y} = F sin 21

        Cos 21 = Fₓ / F

        Fₓ = F cos 21

The friction force equation is

         fr = μ N

We substitute

         F cos 21 - μ N = m a

         N = mg + F sin 21

        F cos 21 - μ (mg + F sin 21) = m a

        a = F cos 21 / m - μ (g + F / m sin 21)

Let's calculate

       a = 322/33 cos 21 - 0.62 (9.8 + 322/33 sin 21)

       a = 9,109 - 8,244

       a = 0.865 m / s²

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s