A 7.80 kg particle starts from rest and moves a distance of 8.90 m in 2.60 s under the action of a single, constant force. Find the magnitude of the force.
vf=at solve for a
f=ma=m*vt/time
To find the magnitude of the force acting on the particle, you need to apply Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.
So, let's start by finding the acceleration of the particle. We can use the kinematic equation:
\[ s = ut + \frac{1}{2}at^2 \]
Where:
s = distance traveled (8.90 m)
u = initial velocity (0 m/s since the particle starts from rest)
a = acceleration
t = time (2.60 s)
Rearranging the equation to solve for acceleration:
\[ a = \frac{2(s - ut)}{t^2} \]
Substituting the given values:
\[ a = \frac{2(8.90 \, \text{m} - 0 \, \text{m/s} \times 2.60 \, \text{s})}{(2.60 \, \text{s})^2} \]
Simplifying:
\[ a = \frac{2(8.90 \, \text{m})}{6.76 \, \text{s}^2} \]
\[ a = \frac{17.80 \, \text{m}}{6.76 \, \text{s}^2} \]
\[ a = 2.625 \, \text{m/s}^2 \]
Now, we can use Newton's second law to find the magnitude of the force:
\[ F = ma \]
Substituting the given mass (7.80 kg) and calculated acceleration (2.625 m/s^2):
\[ F = (7.80 \, \text{kg}) \times (2.625 \, \text{m/s}^2) \]
\[ F = 20.475 \, \text{N} \]
Therefore, the magnitude of the force acting on the particle is 20.475 N.