Force and Kinematics -- Accelerating a 10kg box vertically

AI Thread Summary
To accelerate a 10kg box vertically to a velocity of 5 m/s in 1 second, the upward force must exceed the downward gravitational force of approximately 98 Newtons (F = mg, where g is 9.8 m/s²). The required upward acceleration is calculated as 5 m/s², which can be derived from the formula (Vf - Vi)/t. According to Newton's second law, the net force acting on the box is the sum of the gravitational force and the lifting force, expressed as sigmaF = ma. This means the total upward force must equal the gravitational force plus the force needed for the desired acceleration. Understanding these forces and calculations is crucial for determining the necessary lifting force.
theanswer2physicsisu
Messages
7
Reaction score
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
What forces act on the box? Which ones do you know? What acceleration is needed? What does Newton’s second law tell you?
 
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
 
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 ...
 
Poster has been reminded that they need to show their efforts on their own schoolwork/homework problems
Orodruin said:
... and therefore ...
Therefore what
 
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.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Trying to understand the logic behind adding vectors with an angle between them'
My initial calculation was to subtract V1 from V2 to show that from the perspective of the second aircraft the first one is -300km/h. So i checked with ChatGPT and it said I cant just subtract them because I have an angle between them. So I dont understand the reasoning of it. Like why should a velocity be dependent on an angle? I was thinking about how it would look like if the planes where parallel to each other, and then how it look like if one is turning away and I dont see it. Since...
Back
Top