A shopper in a supermarket pushes a loaded cart with a horizontal force of 80.5N. If the cart has a mass of 33.8kg, how far will it move in 1.31s, starting from rest? The coefficient of friction is 0.12
weight = normal force = m g = 33.8 * 9.81
= 332 N
friction force = 0.12 * 332 = 39.8 N
so net force forward = 80.5 - 39.8
= 40.7 N
a = 40.7/33.8
d = (1/2) a t^2
To find the distance the cart will move, we need to use the equation of motion for an object with constant acceleration:
d = v₀t + 0.5at²
Where:
- d is the distance traveled
- v₀ is the initial velocity (0 in this case since the cart starts from rest)
- t is the time taken
- a is the acceleration
First, we need to find the acceleration of the cart. The net force acting on the cart can be determined using Newton's second law:
F_net = ma
Since the cart is being pushed horizontally, the only horizontal force acting on it is the applied force minus the force of friction:
F_net = F_applied - F_friction
The force of friction can be calculated using the formula:
F_friction = μN
Where:
- μ is the coefficient of friction
- N is the normal force
The normal force can be determined using the equation:
N = mg
Where:
- m is the mass of the cart
- g is the acceleration due to gravity (approximately 9.8 m/s²)
Substituting these values into the equation, we have:
F_friction = μmg
Therefore, the net force can be rewritten as:
F_net = F_applied - μmg
Now we can calculate the acceleration:
a = F_net / m
= (F_applied - μmg) / m
Given that F_applied = 80.5 N, m = 33.8 kg, and μ = 0.12, we can plug in these values to find the acceleration.
Once we have the acceleration, we can use it in the equation of motion to find the distance traveled. Plugging in the values of v₀ = 0, t = 1.31 s, and a from the previous calculation, we can solve for d.