A dump truck, whose bed is made of steel, holds an old steel watering trough. The bed of the truck is slowly raised until the trough begins to slide. For dry steel to steel μs= 0.80, μk= 0.60. What is the acceleration of the trough as it slides down the truck bed? Express your answer with the appropriate units.

a = 1,538 m / s²

Explanation:

Let's use Newton's second law, let's set a reference system where the x-axis is parallel to the sloping floor of the truck and the positive direction is in the direction of movement of the trough, for this case the weight is the only force to decompose

          sin θ = Wₓx / W

          cos θ = W_y / W

          Wₓ = W sin θ

          W_y = W cos θ

Y axis

          N -W_y = 0

          N = mg cos θ

X axis

          Wₓ - fr = m a

the friction force has the expression

          fr = μ N

There are values ​​of the friction coefficient (μ_s) one for when the movement has not started and it takes a smaller value for when the bodies are moving.

In this case we first find the angle for which the movement begins, in this part we use the static coefficient and the acceleration is zero

             Wₓ - μ_s N = 0

             m g sin θ = μ_s mg cos θ

             tan θ = μ_s

             θ = tan⁻¹ μ_s

we calculate

              θ = tan⁻¹ 0.8

              θ = 38.7º

For this angle, how the trough begins to move, the coefficient is reduced to the dynamics coefficient (μ_k)  and the acceleration is different from zero.

we substitute

          mg sin θ - μ_k mg cos θ = m a

          a = g (sin θ - μ_k cos θ)

let's calculate

          a = 9.8 (sin 38.7 - 0.6 cos 38.7)

          a = 1,538 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