Question

An object with mass 100 kg moved in outer space. When it was at location <8, -30, -4> its speed was 5.5 m/s. A single constant force <220, 460, -200> N acted on the object while the object moved from location <8, -30, -4> m to location <14, -21, -7> m. Then a different single constant force <100, 260, 210> N acted on the object while the object moved from location <14, -21, -7> m to location <17, -27, -3> m. What is the speed of the object at this final location

Answer 1

Answer:

v = ( 6.41 i^ + 8.43 j^ + 2.63 k^ ) m/s

Explanation:

We can solve this problem using the kinematic relations, we have a three-dimensional movement, but we can work as three one-dimensional movements where the only parameter in common is time (a scalar).

X axis.

They indicate the initial position x = 8 m, its initial velocity v₀ = 5.5 m / s, the force Fx₁ = 220 N x₁ = 14 m, now the force changes to Fx₂ = 100 N up to the point xf = 17 m. The final speed is wondered.

As this movement is in three dimensions we must find the projection of the initial velocity in each axis, for this we can use trigonometry

the angle fi is with respect to the in z and the angle theta with respect to the x axis.

               sin φ = z / r

                Cos φ = r_p / r

               z = r sin φ

               r_p = r cos φ

the modulus of the vector r can be found with the Pythagorean theorem

               r² = (x-x₀) ² + (y-y₀) ² + (z-z₀) ²

               r² = (14-8) 2 + (-21 + 30) 2+ (-7 +4) 2

               r = √126

               r = 11.23 m

Let's find the angle with respect to the z axis (φfi)

                φ = sin⁻¹ z / r

                φ = sin⁻¹ ( $\frac{-7+4}{11.23}$ )

                φ = 15.5º

Let's find the projection of the position vector (r_p)

                r_p = r cos φ

                r_p = 11.23 cos 15.5

                r_p = 10.82 m

This vector is in the xy plane, so we can use trigonometry to find the angle with respect to the x axis.

                 cos θ = x / r_p

                 θ = cos⁻¹ x / r_p

                 θ = cos⁻¹ ( $\frac{14-8}{10.82}$)  

                 θ = 56.3º

taking the angles we can decompose the initial velocity.

               sin φ = v_z / v₀

               cos φ = v_p / v₀

               v_z = v₀ sin φ

               v_z = 5.5 sin 15.5 = 1.47 m / z

               v_p = vo cos φ

               v_p = 5.5 cos 15.5 = 5.30 m / s

                 

               cos θ = vₓ / v_p

                sin θ = v_y / v_p

                vₓ = v_p cos θ

                v_y = v_p sin θ

                vₓ = 5.30 cos 56.3 = 2.94 m / s

                v_y = 5.30 sin 56.3 = 4.41 m / s

 

                 

we already have the components of the initial velocity

                v₀ = (2.94 i ^ + 4.41 j ^ + 1.47 k ^) m / s

let's find the acceleration on this axis (ax1) using Newton's second law

                Fₓx = m aₓ₁

                aₓ₁ = Fₓ / m

                aₓ₁ = 220/100

                aₓ₁ = 2.20 m / s²

Let's look for the velocity at the end of this interval (vx1)

Let's be careful if the initial velocity and they relate it has the same sense it must be added, but if the velocity and acceleration have the opposite direction it must be subtracted.

                 vₓ₁² = v₀ₓ² + 2 aₓ₁ (x₁-x₀)

                 

let's calculate

                 vₓ₁² = 2.94² + 2 2.20 (14-8)

                 vₓ₁ = √35.04

                 vₓ₁ = 5.92 m / s

to the second interval

they relate it to xf

                   aₓ₂ = Fₓ₂ / m

                   aₓ₂ = 100/100

                   aₓ₂ = 1 m / s²

final speed

                    v_{xf}²  = vₓ₁² + 2 aₓ₂ (x_f- x₁)

                    v_{xf}² = 5.92² + 2 1 (17-14)

                    v_{xf} =√41.05

                    v_{xf} = 6.41 m / s

We carry out the same calculation for each of the other axes.

Axis y

acceleration (a_{y1})

                      a_{y1} = F_y / m

                      a_{y1} = 460/100

                      a_[y1} = 4.60 m / s²

the velocity at the end of the interval (v_{y1})

                      v_{y1}² = v_{oy}² + 2 a_{y1{ (y₁ -y₀)

                      v_{y1}2 = 4.41² + 2 4.60 (-21 + 30)

                      v_{y1} = √102.25

                       v_{y1} = 10.11 m / s

second interval

acceleration (a_{y2})

                      a_{y2} = F_{y2} / m

                      a_{y2} = 260/100

                      a_{y2} = 2.60 m / s2

final speed

                     v_{yf}² = v_{y1}² + 2 a_{y2} (y₂ -y₁)

                     v_{yf}² = 10.11² + 2 2.60 (-27 + 21)

                      v_{yf} = √ 71.01

                      v_{yf} = 8.43 m / s

here there is an inconsistency in the problem if the body is at y₁ = -27m and passes the position y_f = -21 m with the relationship it must be contrary to the velocity

z axis

 

first interval, relate (a_{z1})

                      a_{z1} = F_{z1} / m

                      a_{z1} = -200/100

                      a_{z1} = -2 m / s

the negative sign indicates that the acceleration is the negative direction of the z axis

the speed at the end of the interval

                    v_{z1}² = v_{zo)² + 2 a_{z1} (z₁-z₀)

                    v_{z1}² = 1.47² + 2 (-2) (-7 + 4)

                    v_{z1} = √14.16

                    v_{z1} = -3.76 m / s

second interval, acceleration (a_{z2})

                    a_{z2} = F_{z2} / m

                    a_{z2} = 210/100

                    a_{z2} = 2.10 m / s2

final speed

                    v_{fz}² = v_{z1}² - 2 a_{z2} | z_f-z₁)

                    v_{fz}² = 3.14² - 2 2.10 (-3 + 7)

                     v_{fz} = √6.94

                     v_{fz} = 2.63 m / s

speed is     v = ( 6.41 i^ + 8.43 j^ + 2.63 k^ ) m/s