Problem::
A 4.80 kg block located on a horizontal frictionless floor is pulled by a cord that exerts a force F=13.5 N at an angle q= 17.5° above the horizontal. What is the speed of the block 3.90 seconds after it starts moving?
What I've got::
Using sin and cos, I've got that Viy is 4.059 and Vix is 12.875.
So, using 12.875 for my initial velocity and the formula Vf=Vi+at, I get Vf=12.875+(2.81)(3.9), which gives me Vf=23.84, but that isn't right. Can someone show me what I'm doing wrong?
Vi is zero.
12.875 is the pulling force
a=pullingforce/m=12.875/4.80=your acceleration
Vf=a*3.9seconds
To solve this problem, you need to break down the force F into its horizontal and vertical components. The horizontal component of the force will cause acceleration, while the vertical component will not affect the speed of the block.
Let's start by finding the horizontal and vertical components of the force F:
F_horizontal = F * cos(q)
F_vertical = F * sin(q)
Given:
F = 13.5 N
q = 17.5°
F_horizontal = 13.5 * cos(17.5°)
F_horizontal = 12.874 N
F_vertical = 13.5 * sin(17.5°)
F_vertical = 4.0589 N
Now we can find the acceleration using Newton's second law:
F_horizontal = m * a
12.874 = 4.8 * a
a = 12.874 / 4.8
a = 2.682 N/kg
Since the floor is frictionless, the horizontal acceleration will remain constant.
Next, we can use the equation of motion to find the final velocity of the block after 3.90 seconds:
Vf = Vi + a * t
Here, the initial velocity Vi is given as 0 because the block starts from rest:
Vf = 0 + 2.682 * 3.9
Vf = 10.4558 m/s
So the speed of the block 3.90 seconds after it starts moving is approximately 10.46 m/s.
In your calculation, you mistakenly used the initial velocity in the horizontal direction (Vix) as 12.875 instead of using 0 because the block starts from rest. That is why you obtained an incorrect result.