A 92.7 N grocery cart is pushed 13.7 m along
an aisle by a shopper who exerts a constant
horizontal force of 33.4 N.
The acceleration of gravity is 9.81 m/s
2
.
If all frictional forces are neglected and the
cart starts from rest, what is the grocery cart’s
final speed?
Answer in units of m/s
To find the final speed of the grocery cart, we can use the equation that relates acceleration, force, and mass:
F = ma
Here, F is the applied force, m is the mass of the cart, and a is the acceleration.
The mass of the cart can be determined using the equation:
F = mg
Where g is the acceleration due to gravity.
Rearranging the equation for mass, we have:
m = F / g
Substituting the given values, we have:
m = 92.7 N / 9.81 m/s^2
m ≈ 9.45 kg
Now, we can calculate the acceleration using the equation:
a = F / m
Substituting the given values, we have:
a = 33.4 N / 9.45 kg
a ≈ 3.53 m/s^2
Since the cart starts from rest, its initial velocity is 0 m/s. We can use the equation that relates acceleration, final velocity, initial velocity, and displacement:
v^2 = u^2 + 2as
where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement.
Rearranging the equation, we have:
v^2 = u^2 + 2as
v^2 = 0^2 + 2(3.53 m/s^2)(13.7 m)
v^2 ≈ 96.672 m^2/s^2
Taking the square root of both sides to find the final velocity, we get:
v ≈ √(96.672 m^2/s^2)
v ≈ 9.83 m/s
Therefore, the grocery cart's final speed is approximately 9.83 m/s.