Force and Kinematics -- Accelerating a 10kg box vertically

In summary: From this, you can calculate the force that is needed to lift the object up at the given acceleration.
  • #1
theanswer2physicsisu
7
0
Homework Statement
How much force is required to lift a 10 kg box such that it is accelerated from rest to a velocity of 5 m/s within 1 second?
Relevant Equations
F = ma ; v/t = a; Kinematics; Force Equations
I realize that there is a downward force of gravity weighing the object toward earth’s surface, equaling F = mg (downward). The upward force would have to be something at least as much as the downward force in order to lift the object up ”such that it is accelerated from rest to a velocity of 5 m/s within 1 second” as stated in the question stem. The remaining question is how much more Newtons (N) of force is required in order to lift the object up at that acceleration in that amount of time?
 
Physics news on Phys.org
  • #2
What forces act on the box? Which ones do you know? What acceleration is needed? What does Newton’s second law tell you?
 
  • #3
Orodruin said:
What forces act on the box? Which ones do you know? What acceleration is needed? What does Newton’s second law tell you?
Gravitation force downward, lifting force upward, gravitation acceleration is around 9.8 m/s^2, and the acceleration might be calculated via: (Vf-Vi)/t = (5m/s-0m/s)/1s = 5m/s^2. This acceleration of 5 m/s^2 may be plugged into the force equation of F = ma. Newton’s 2nd law tells us a net force acting on an object causes change in object’s motion inversely proportional to mass and directly proportional to the net force acting on the object, sigmaF= ma
 
  • #4
theanswer2physicsisu said:
Gravitation force downward, lifting force upward, gravitation acceleration is around 9.8 m/s^2, and the acceleration might be calculated via: (Vf-Vi)/t = (5m/s-0m/s)/1s = 5m/s^2. This acceleration of 5 m/s^2 may be plugged into the force equation of F = ma. Newton’s 2nd law tells us a net force acting on an object causes change in object’s motion inversely proportional to mass and directly proportional to the net force acting on the object, sigmaF= ma
... and therefore ...
 
  • #5
Poster has been reminded that they need to show their efforts on their own schoolwork/homework problems
Orodruin said:
... and therefore ...
Therefore what
 
  • #6
With that information you just need to piece the things together. You have stated which forces act on the box, you know what the gravitational force is, you know what the mass is, you know what the acceleration is and you know that ma is the sum of the forces.
 

1. What is the formula for calculating force?

The formula for calculating force is F = m x a, where F is force in Newtons (N), m is mass in kilograms (kg), and a is acceleration in meters per second squared (m/s^2).

2. How do you calculate the acceleration of a 10kg box?

To calculate the acceleration of a 10kg box, you need to know the force acting on the box. Once you have the force, you can use the formula a = F/m, where a is acceleration, F is force, and m is mass. For example, if the force acting on the box is 50 N, the acceleration would be 50 N / 10 kg = 5 m/s^2.

3. What is the unit of force?

The unit of force is Newton (N). Other units of force include pound-force (lbf) and kilopond (kp).

4. How does the mass of an object affect its acceleration?

The mass of an object has a direct relationship with its acceleration. This means that as the mass increases, the acceleration decreases. This can be seen in the formula a = F/m, where a is inversely proportional to m. In other words, as mass increases, acceleration decreases, and vice versa.

5. Can an object have a negative acceleration?

Yes, an object can have a negative acceleration. This means that the object is slowing down in the direction of motion. Negative acceleration is also known as deceleration or retardation. It can be calculated using the same formula as acceleration, but with a negative sign in front of the value.

Similar threads

  • Introductory Physics Homework Help
Replies
5
Views
799
  • Introductory Physics Homework Help
Replies
14
Views
355
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
2
Replies
38
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
373
  • Introductory Physics Homework Help
Replies
17
Views
950
  • Introductory Physics Homework Help
Replies
1
Views
909
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
15
Views
2K
Back
Top