Two forces, and act on a particle of mass 1.50 kg that is initially at rest at c

Question

Two forces,

and

act on a particle of mass 1.50 kg that is initially at rest at coordinates

(?1.80 m, +3.75 m).

Besides the gravitational force, a 2.30-kg object is subjected to one other constant force. The object starts from rest and in 1.20 s experiences a displacement of (5.10 r = m (d) What are the coordinates of the particle at t = 12.0 s? x = m y = m v = m/s (b) In what direction is the particle moving at t = 12.0 s? A degree counterclockwise from the +x-axis (c) What displacement does the particle undergo during the first 12.0 s? ? ) N, act on a particle of mass 1.50 kg that is initially at rest at coordinates (?1.80 m, +3.75 m). (a) What are the components of the particle's velocity at t = 12.0 s? ? 4.30 F 2 = (?3.30 ) N and ? 5.55 F 1 = (?6.20 is the upward vertical direction. Determine the other force. Two forces,

Explanation / Answer

So, we assume constant acceleration in both the "i" and "j" directions.

In the "i" direction find the acceleration:
sf = 5.10 m
Assume si = 0 m
vi = 0 m
t = 1.20 s

Equation of motion:
sf = si + vi*t + 1/2*a*t^2

With si and vi = 0
sf = 1/2*a*t^2
a = 2*sf / t^2

Plug in numbers:
a = 2*(5.10 m) / (1.20 s)^2
a = 7.083 m/s^2

Definition of force (constant mass)
F = m*a
F = 2.30 kg * 7.083 m/s^2
F = 16.2916 N

This is our "i" component

For the "j" direction:
hf = -3.30 m
assume hi = 0 m
vi = 0 m/s
t = 1.20 s

Equation of motion
hf = hi + vi*t + 1/2*a*t^2

With hi and vi = 0:
hf = 1/2*a*t^2
a = 2*hf / t^2
a = 2*(-3.30 m) / (1.20 s)^2
a = -4.58 m/s^2

However, since gravity is also acting on the object in the vertical direction we have to separate out the acceleration due to the force. The acceleration we found is the acceleration from the force and from gravity. There are two ways, you can just assume that the acceleration due to the force is "a-g" or you can use a force balance. I'll use the latter. Note: The "a-g" assumes up is positive, so g is negative and you'll end up with "a+g" using the value of 9.81 m/s^2 for g instead of -9.81 m/s^2 using the "a-g".

Sum of the forces in the vertical direction (up is positive) = m*a = F - m*g
F = m*(a+g)

Plug in numbers
F = 2.30 kg * (-4.58 m/s^2 + 9.81 m/s^2)
F = 11.992 N

So, your force vector is
F = 16.2916 i + 11.992 j N

You can determine the magnitude and direction yourself if needed.

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