A shopper in a supermarket pushes a loaded 36kg cart with a horizontal force of 12N. (Ignore Friction) (a) How far will the cart move in 3.5s, starting from rest? m (b) How far will the cart move in 3.5s if the shopper places an 75N child in the cart before pushing? m
a = F/m = 12/36 = 1/3 m/s^2
d = (1/2) a t^2 = (1/2)(1/3) (3.5)^2
for part b, use 36 + 75 instead of 36
a = 12/111
Well, isn't it cart-tastic to have a loaded question like this? Let's calculate the distance this shopper will cover!
(a) Since the cart is initially at rest, we can use Newton's second law: F = ma, where F is the force applied, m is the mass of the cart, and a is the acceleration.
We can find the acceleration with a = F/m. Plugging in the values, we have a = 12N / 36kg = 0.333 m/s².
Now, we can use the equation for distance covered with constant acceleration: d = (1/2)at², where d is the distance, a is the acceleration, and t is the time.
Plugging in the values, we get d = (1/2) * 0.333 m/s² * (3.5s)². Crunching the numbers, we find that the cart will move approximately 2.704 meters.
(b) Now, the shopper decides to turn this shopping trip into a little joy ride for a 75N child. To calculate the total force applied, we need to add the force exerted by the shopper (12N) to the weight of the child (75N). So, the total force is 12N + 75N = 87N.
Following the same steps as before, we obtain an acceleration of a = F/m = 87N / 36kg = 2.417 m/s².
Using the equation d = (1/2)at² with the new acceleration, we find that the cart will move approximately 12.801 meters.
So, in conclusion, without the child, the cart will move about 2.704 meters, but with the child, it will move around 12.801 meters. That's quite the difference! Time to get that shopping trip on a roll!
To calculate the distance the cart will move in each scenario, we can use the equation of motion:
distance = (1/2) * acceleration * time^2
First, let's calculate the acceleration in each case.
(a) When the shopper pushes the cart with a horizontal force of 12N:
In this case, we need to calculate the acceleration using Newton's second law:
force = mass * acceleration
The mass of the cart is 36kg, and the force applied is 12N.
12N = 36kg * acceleration
Now we can solve for acceleration:
acceleration = 12N / 36kg
(b) When the shopper places a 75N child in the cart before pushing:
In this case, the total force applied to the cart is the sum of the force applied by the shopper and the force due to the child's weight. So,
total force = force by shopper + force due to child's weight
The force by the shopper is 12N, and the force due to the child's weight is 75N.
total force = 12N + 75N = 87N
We can now calculate the acceleration similarly to the previous case:
87N = 36kg * acceleration
acceleration = 87N / 36kg
Now that we have the acceleration values for both cases, we can calculate the distances.
(a) Distance when pushing with a force of 12N:
distance = (1/2) * acceleration * time^2
distance = (1/2) * (12N / 36kg) * (3.5s)^2
(b) Distance when pushing with a force of 87N (with a 75N child in the cart):
distance = (1/2) * acceleration * time^2
distance = (1/2) * (87N / 36kg) * (3.5s)^2
Now we can calculate the distances using the given values.
To find the distance the cart will move in both scenarios, we need to use Newton's second law of motion, which states that force is equal to mass multiplied by acceleration (F = m * a). Since the cart is moving horizontally without any friction, the only force acting on it is the force applied by the shopper.
Let's break down each scenario separately:
(a) When the cart is pushed with a force of 12N, the net force acting on the cart is simply the force applied by the shopper. We can use this net force to calculate the acceleration of the cart by rearranging the formula:
F = m * a
12N = 36kg * a
Now we can solve for acceleration:
a = 12N / 36kg
a = 0.33 m/s^2
Since the cart starts from rest, its initial velocity (v0) is 0 m/s. We can use the kinematic equation that relates distance, time, acceleration, and initial velocity to find the distance (d):
d = v0 * t + 0.5 * a * t^2
Plugging in the values:
d = 0 * 3.5s + 0.5 * 0.33 m/s^2 * (3.5s)^2
d = 0 + 0.5 * 0.33 m/s^2 * 12.25s^2
d = 0 + 2.40825 m
d = 2.41 m
Therefore, the cart will move approximately 2.41 meters in 3.5 seconds when the shopper pushes it with a force of 12N.
(b) When the child is placed in the cart, there is an additional force acting on the cart due to the weight of the child. To calculate the net force, we need to consider both the force applied by the shopper and the weight of the child.
The weight of the child is equal to the mass of the child multiplied by the acceleration due to gravity (w = m * g). We can calculate the weight of the child using this formula:
w = 75N
Now let's calculate the net force acting on the cart:
Net Force = Force by shopper - Weight of child
Net Force = 12N - 75N
Net Force = -63N
Since the net force is negative (-63N), the cart will experience deceleration in this scenario. We can calculate the acceleration using Newton's second law:
F = m * a
-63N = 36kg * a
Now we can solve for acceleration:
a = -63N / 36kg
a = -1.75 m/s^2
With the same initial velocity and time as in the previous scenario, we can use the kinematic equation to find the distance (d):
d = v0 * t + 0.5 * a * t^2
Plugging in the values:
d = 0 * 3.5s + 0.5 * -1.75 m/s^2 * (3.5s)^2
d = 0 + 0.5 * -1.75 m/s^2 * 12.25s^2
d = 0 + -10.68 m
d = -10.68 m
Since distance cannot be negative, we take the magnitude of the distance, resulting in approximately 10.68 meters.
Therefore, the cart will move approximately 10.68 meters in 3.5 seconds when the shopper places a 75N child in the cart before pushing.