A shopper in a supermarket pushes a loaded cart with a horizontal force of 12 N. The cart has a mass of 33 kg.
(a) How far will it move in 7.0 s, starting from rest? (Ignore friction.)
m
(b) How far will it move in 7.0 s if the shopper places her 30 N child in the cart before she begins to push it?
m
To determine the distance the cart will move, we need to use the equation of motion. The equation we will use is:
d = 0.5 * a * t^2
Where:
d is the distance traveled
a is the acceleration
t is the time
(a) First, let's find the acceleration using Newton's second law of motion:
F = m * a
Where:
F is the force applied
m is the mass of the object
a is the acceleration
In this case,
F = 12 N (the applied force)
m = 33 kg (mass of the cart)
Substituting the known values into the equation, we can find the acceleration:
12 N = 33 kg * a
Solving for a:
a = 12 N / 33 kg
a ≈ 0.364 m/s^2
Next, we can substitute the acceleration and time into the equation of motion to find the distance traveled:
d = 0.5 * 0.364 m/s^2 * (7.0 s)^2
d ≈ 0.5 * 0.364 m/s^2 * 49 s^2
d ≈ 8.882 m
So, the cart will move approximately 8.882 meters in 7.0 seconds.
(b) Now, let's consider the case where the shopper places her 30 N child in the cart before pushing it. In this case, the total force applied to the cart will be the sum of the force applied by the shopper and the force applied by the child:
F_total = F_shopper + F_child
F_total = 12 N + 30 N
F_total = 42 N
Using Newton's second law of motion again, we can find the new acceleration:
42 N = 33 kg * a
Solving for a:
a = 42 N / 33 kg
a ≈ 1.273 m/s^2
Substituting the new acceleration and time into the equation of motion:
d = 0.5 * 1.273 m/s^2 * (7.0 s)^2
d ≈ 0.5 * 1.273 m/s^2 * 49 s^2
d ≈ 30.143 m
Therefore, the cart will move approximately 30.143 meters in 7.0 seconds when the shopper places her 30 N child in the cart.
No friction?
f=ma so a=f/m
a. distance=1/2 a t^2
b. again, just change mass.