A shopper in a supermarket pushes a loaded cart with a horizontal force of 13 N. The cart has a mass of 32 kg.
(a) How far will it move in 4.0 s, starting from rest? (m) (Ignore friction.)
(b) How far will it move in 4.0 s if the shopper places his 30 N child in the cart before he begins to push it? (m)
0.60 m
To solve both parts of the problem, we can use Newton's second law, which states that the net force acting on an object is equal to its mass times its acceleration.
(a) To find the distance the cart will move in 4.0 seconds when starting from rest, we need to calculate the acceleration of the cart first. We can use the equation F = ma, where F is the applied force, m is the mass of the cart, and a is the acceleration.
Given:
Force applied (F) = 13 N
Mass of the cart (m) = 32 kg
Using the equation F = ma, we can rearrange it to solve for acceleration (a):
a = F / m
Substituting the given values, we have:
a = 13 N / 32 kg
Now, we can use the kinematic equation s = ut + (1/2)at^2 to find the distance traveled (s). The initial velocity (u) of the cart is zero because it starts from rest, and the time (t) is 4.0 seconds.
Rearranging the equation, we get:
s = (1/2)at^2
Substituting the values, we have:
s = (1/2)*(13 N / 32 kg)*(4.0 s)^2
Now we can solve for s:
s = (1/2)*(13 N / 32 kg)*(16.0 s^2)
s = (1/64)*(13 N)*(16.0 s^2)
s = (13 N)*(0.25 s^2)
s = 3.25 m
Therefore, the cart will move 3.25 meters in 4.0 seconds when starting from rest.
(b) In this case, we need to consider the additional mass of the child when calculating the distance traveled. The mass of the cart will now include the child's mass.
Given:
Force applied (F) = 13 N
Mass of the cart (m) = 32 kg
Force of the child's weight (F_child) = 30 N
Adding the weight of the child to the cart, the total mass (m_total) is:
m_total = m_cart + m_child
Substituting the values:
m_total = 32 kg + (30 N / 9.8 m/s^2) kg
Using the equation F = ma, we can find the new acceleration (a):
a = F_total / m_total
Substituting the values:
a = (13 N) / (32 kg + (30 N / 9.8 m/s^2) kg)
Now, using the kinematic equation s = ut + (1/2)at^2, we can solve for the distance (s) with the new acceleration, initial velocity of zero, and the same time (t) of 4.0 seconds.
Rearranging the equation, we get:
s = (1/2)at^2
Substituting the values, we have:
s = (1/2)*(a)*t^2
Now we can solve for s using the new acceleration value:
s = (1/2)*[(13 N) / (32 kg + (30 N / 9.8 m/s^2) kg)]*(4.0 s)^2
Simplifying the expression, we find:
s ≈ 2.13 meters
Therefore, the cart will move approximately 2.13 meters in 4.0 seconds when the shopper places his 30 N child in the cart before pushing it.