Three forces acting on an object are given by F1 = ( - 1.85i + 2.50j) N, F2 = ( - 5.10i + 3.25j ) N, and

F3 = ( - 45.0i + 48.0j ) N. The object experiences an acceleration of magnitude 3.75 m/s2.

(a) What is the direction of the acceleration?
_________° (from the positive x axis)

(b) What is the mass of the object?
_________ kg

(c) If the object is initially at rest, what is its speed after 17.0 s?
_________ m/s

(d) What are the velocity components of the object after 17.0 s?
( ________ i + _________ j ) m/s

a) direction of acceleration in the same way as resultant force, hence

Fr=F1+F2+F3=(-1.85-5.10-45.0)i+(2.50+3.25+48.0)j= -51.95i+53.75j [N]
θ=tanˉ¹(53.75/-51.95)=134°.
b) F=ma --- m=F/a --- (√((-51.95)^2+53.75^2))/3.75=19.9kg
c)Vf=Vi+at=0+3.75*17=63.75m/s
d)F=ma --- a=F/m ---āx=F̅x/m --āy=F̅y/m
ā=āx + āy
V̅f=V̅i+āt=0+((-51.95/19.9)i+(53.75/19.5)j)*10
V̅f=((-51.95/19.9)*10)i+(53.7/19.5)*10)j
V̅f= -26.1i+27.5j

This is an exercise in adding vectors. The ratio of components in vector sum will provide the fdirection information you need. The mass is the ratio of the force magnitude to the acceleration.

For (c), assume the initial speed is zero and multiply the acceleration and time interval. For (d), multiply the vector acceleration by the time.

Show your work if further assistance is needed.

To find the direction of the acceleration, we need to calculate the net force acting on the object. The net force is the vector sum of the individual forces acting on the object:

Net Force = F1 + F2 + F3

Given:

F1 = (-1.85i + 2.50j) N
F2 = (-5.10i + 3.25j) N
F3 = (-45.0i + 48.0j) N

Let's calculate the net force:

Net Force = (-1.85i + 2.50j) N + (-5.10i + 3.25j) N + (-45.0i + 48.0j) N

Simplifying, we get:

Net Force = (-51.95i + 53.75j) N

The magnitude of the acceleration is given as 3.75 m/s^2. We know that the acceleration is related to the net force by Newton's Second Law:

Net Force = mass × acceleration

Let's solve for the magnitude of the mass:

mass = (magnitude of Net Force) / (magnitude of acceleration)

mass = √[(-51.95)^2 + (53.75)^2] N / 3.75 m/s^2

Calculating the magnitude of the mass, we find:

mass = √[2693.1025 + 2885.5625] N / 3.75 m/s^2

mass = √5578.665 N / 3.75 m/s^2

mass = 74.70 kg (rounded to two decimal places)

Now, let's move on to the next question.

To find the speed of the object after 17.0 s, we need to use the equations of motion.

The acceleration of the object is constant, so we can use the equation:

v = u + at

where:
v = final velocity
u = initial velocity (zero since the object is initially at rest)
a = acceleration
t = time

Plugging in the values:

v = 0 + (3.75 m/s^2)(17.0 s)

Calculating, we find:

v = 63.75 m/s

Therefore, the speed of the object after 17.0 s is 63.75 m/s.

Finally, let's find the velocity components of the object after 17.0 s.

The velocity is given by:

v = vi + vj

where:
vi = velocity component in the i-direction
vj = velocity component in the j-direction

We can use the equation:

vi = u + at
vj = v0 + at

Given:
u = 0 (initial velocity)
a = 3.75 m/s^2 (acceleration)
t = 17.0 s (time)

vi = 0 + (3.75 m/s^2)(17.0 s)
vj = 0 + (3.75 m/s^2)(17.0 s)

Calculating, we find:

vi = 63.75 m/s
vj = 63.75 m/s

Therefore, the velocity components of the object after 17.0 s are (63.75i + 63.75j) m/s.

To summarize the answers:

(a) The direction of the acceleration is given by the angle between the net force vector and the positive x-axis.
(b) The mass of the object is approximately 74.70 kg.
(c) The speed of the object after 17.0 s is approximately 63.75 m/s.
(d) The velocity components of the object after 17.0 s are (63.75i + 63.75j) m/s.

To find the direction of the acceleration, we can use the three forces acting on the object.

Step 1: Find the net force acting on the object.
The net force is the vector sum of the three forces:
Fnet = F1 + F2 + F3

Step 2: Use Newton's second law to find the magnitude and direction of the acceleration.
Newton's second law states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration:
Fnet = m * a

(a) Find the direction of the acceleration:
The direction of the acceleration can be determined by finding the angle between the net force vector and the positive x-axis.

Step 3: Find the magnitude of the acceleration:
Using the given information:

F1 = -1.85i + 2.50j N
F2 = -5.10i + 3.25j N
F3 = -45.0i + 48.0j N

Fnet = F1 + F2 + F3

Fnet = (-1.85i + 2.50j) + (-5.10i + 3.25j) + (-45.0i + 48.0j)
= -1.85i + 2.50j - 5.10i + 3.25j - 45.0i + 48.0j
= (-1.85 - 5.10 - 45.0)i + (2.50 + 3.25 + 48.0)j
= -52.95i + 53.75j N

The magnitude of the acceleration can be found using the formula:
|Fnet| = m * a

|Fnet| = √((-52.95)^2 + (53.75)^2) N
|Fnet| = √(2795.2025 + 2885.5625) N
|Fnet| = √(5680.765) N
|Fnet| = 75.37 N

Therefore, the acceleration of the object has a magnitude of 75.37 m/s^2.

(b) Now we can find the mass of the object:
Using Newton's second law:

|Fnet| = m * a

75.37 N = m * 3.75 m/s^2

Solving for m, we get:

m = 75.37 N / 3.75 m/s^2
m = 20.10 kg

Therefore, the mass of the object is 20.10 kg.

(c) To find the speed of the object after 17.0 s, we need to find the final velocity. Since the object is initially at rest and has been subjected to a constant acceleration, we can use the following equation of motion:

vf = vi + a * t

Where vf is the final velocity, vi is the initial velocity (which is 0 in this case), a is the acceleration, and t is the time.

Using the given information:

vi = 0 m/s
a = 3.75 m/s^2
t = 17.0 s

vf = 0 + 3.75 * 17.0
vf = 63.75 m/s

Therefore, the speed of the object after 17.0 s is 63.75 m/s.

(d) To find the velocity components of the object after 17.0 s, we can use the following equation:

vf = (vf_x)i + (vf_y)j

Where vf_x is the x-component of the final velocity, and vf_y is the y-component of the final velocity.

Using the given information:

vf = 63.75 m/s

We need to find the x and y components of vf.

Since the object is subjected to a constant acceleration over time, we can use the kinematic equation:

vf = vi + a * t

In the x-direction, vi = 0, a = -52.95 (from previous calculations), and t = 17.0 s.

vf_x = 0 + (-52.95) * 17.0
vf_x = -898.65 m/s

In the y-direction, vi = 0, a = 53.75 (from previous calculations), and t = 17.0 s.

vf_y = 0 + 53.75 * 17.0
vf_y = 914.75 m/s

Therefore, the velocity components of the object after 17.0 s are (-898.65 i + 914.75 j) m/s.