F = ma, net force on shopping carts

In summary, the employee is pushing ten shopping carts with an acceleration of 0.05 m/s^2 in a straight line. Each cart has a mass of 26 kg and the ground is level. The net force acting on any one of the carts is 1.3 Newtons. Assuming friction is negligible, the force exerted by the fifth cart on the sixth cart is 6.5 Newtons. The net force on the group of carts is different from the net force on each individual cart, and this can be explained by Newton's second law and the concept of momentum.
  • #1
jehan4141
91
0
In a supermarket parking lot, an employee is pushing ten empty shopping carts, lined up in a straight line. The acceleration of the carts is 0.05 m/s^2. The ground is level, and each cart has a mass of 26 kg.

a. What is the net force acting on anyone of the carts?
b. Assuming friction is negligible, what is the force exerted by the fifth cart on the sixth cart?

Answers:
a. 1.3 Newtons
b. 6.5 Newtons

I most definitely understand the concept behind B and how to reach the answer of 6.5 Newtons.

I know that to get answer A, you do F = ma = 26(0.05) =1.3 Newtons...However, I am confused as to why part A is not 13 Newtons.

Isn't the net force on any single cart going to be 13 Newtons? My reasoning was that there are ten carts total, for a total mass of 260 kilograms.

Total mass = 10(26) = 260 kg

Fnet = total mass x acceleration = 260 x 0.05 = 13 Newtons.

Don't the masses of the carts add up to give the net force on any cart??
 
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  • #2
jehan4141 said:
In a supermarket parking lot, an employee is pushing ten empty shopping carts, lined up in a straight line. The acceleration of the carts is 0.05 m/s^2. The ground is level, and each cart has a mass of 26 kg.

a. What is the net force acting on anyone of the carts?
b. Assuming friction is negligible, what is the force exerted by the fifth cart on the sixth cart?

Answers:
a. 1.3 Newtons
b. 6.5 Newtons

I most definitely understand the concept behind B and how to reach the answer of 6.5 Newtons.

I know that to get answer A, you do F = ma = 26(0.05) =1.3 Newtons...However, I am confused as to why part A is not 13 Newtons.

Isn't the net force on any single cart going to be 13 Newtons? My reasoning was that there are ten carts total, for a total mass of 260 kilograms.

Total mass = 10(26) = 260 kg

Fnet = total mass x acceleration = 260 x 0.05 = 13 Newtons.

Don't the masses of the carts add up to give the net force on any cart??

If you can work out how strongly Cart 5 pushes on Cart 6, try cart 3 on cart 4 and all the way back to the employee pushing Cart 1.
Then consider the difference between the net force on the bunch of carts vs the net force on each of them.
 
  • #3
Newton's second law states that the net force on an object is equal to the rate of change of momentum (F=ma for this case). Does (a) make sense then?
 
  • #4
I do not know what momentum is. I have just started physics a month ago. I've been looking at this problem for 30 minutes and I still do not understand.
 
  • #5
Well, you might want to study out of a proper textbook in that case. (a) is a straight forward application of Newton's second law. Given the mass m=26 kg and the acceleration a=0.05 m/s2, the net force is given by (ma) or 1.3 N. The net force comes out from the forward force due to the previous carts and the reaction force due to the carts in front.
 
  • #6
jehan4141 said:
I do not know what momentum is. I have just started physics a month ago. I've been looking at this problem for 30 minutes and I still do not understand.

Imagine instead you were holding a long pole, with 10 separate strings attached to the 10 carts which were lined up beside each other.

You then accelerate them at 0.05 m/s^2

Can you now see the difference between the total force you apply, as compared to the force on each of the trollies.

In each case you apply a force to the set of trollies, and each trolley is accelerated.
 
  • #7
Pi-bond: No need to be rude...If I could afford a proper textbook, don't you think I would get one? I am 13 years old, unable to get a job just yet, and my mom is a heroine-addicted, single mother working minimum wage. I can't afford a textbook that is $100+... so I rely on borrowing my friend's free internet to learn and when I can get there, I go to the library to use the limited textbooks available...and I can only use these INSIDE the public library during business hours... I get help where I can, and I'm not afraid to ask for help when I need it. EFF THIS SITE. I'm out.
 
  • #8
Thank you PeterO, I think I am just over-thinking and over-analyzing the question. I do that a lot..
 
  • #9
your second example just clicked :D thank you again :)))))) -jehan
 

1. What is the meaning of "F = ma"?

"F = ma" is a mathematical expression known as Newton's second law of motion. It states that the net force applied on an object is equal to its mass multiplied by its acceleration. In simpler terms, it explains how an object's motion is affected by the forces acting upon it.

2. How does "F = ma" apply to shopping carts?

"F = ma" applies to shopping carts because they are objects that have mass and experience forces. When you push a shopping cart, you are applying a force (F) to it, causing it to accelerate (a) in the direction of the force. The mass (m) of the shopping cart also plays a role in determining how much force is needed to accelerate it.

3. What is the net force on a shopping cart?

The net force on a shopping cart is the sum of all the forces acting on it. In other words, it takes into account the forces pushing or pulling in different directions. For example, if you push a shopping cart with a force of 10 Newtons, but there is also a frictional force of 5 Newtons acting against it, the net force on the shopping cart would be 5 Newtons in the direction of the push.

4. How does the net force affect the motion of a shopping cart?

According to Newton's second law, the net force acting on an object determines its acceleration. So, if the net force on a shopping cart is positive, meaning there is a greater force pushing it than any opposing forces, the cart will accelerate in the direction of the force. On the other hand, if the net force is negative, the cart will decelerate or slow down in the opposite direction of the force.

5. Why is it important to consider the net force on shopping carts?

Considering the net force on shopping carts is important because it helps us understand how different forces affect the motion of the cart. It also allows us to calculate the necessary force needed to move or stop the cart, which can be useful in designing more efficient and safe shopping carts. Additionally, being mindful of the net force can help prevent accidents or injuries caused by uncontrolled or excessive motion of the shopping cart.

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