Acceleration upwards and the effect of 'g'

In summary: Only when the body is in free fall, with no other forces acting on it, do we use the value of 9.8 m/s2 for the acceleration due to gravity. In summary, the conversation discusses the concept of acceleration, specifically in relation to gravity and forces. It also includes a question involving a lamp hanging in a descending elevator and how to calculate its mass. The expert clarifies that the acceleration due to gravity is not always subtracted or added in calculations and provides an example to illustrate this.
  • #1
Xenoned
2
0
Hi,

If a statement something like this is given:
"A body is going up with a constant acceleration of 2 m/s^2"

Does it mean that acceleration due to gravity acts on it and we have to subtract 9.8 from 25?

We subtract forces by Newton's laws of motion right?

There is a question like this:
"A lamp hangs vertically from a chord in a descending lift. The lift has a deceleration of 5.2 m/s2 before coming to a halt. If the tension in cord is 30 N, find the mass of the lamp."

We answer it by T=m(g+a) ; a= acceleration/deceleration and get the answer.
but here we use g+a .

why not just acceleration?
Why shouldn't we subtract or add g in case of 1D motion?
 
Last edited:
Physics news on Phys.org
  • #2
Where do you get 25 from?

If a body is accelerating upward at 2 m/s^2, it cannot at the same time be accelerating at 9.8 m/s^2 in the opposite direction.

In the elevator question, you should draw a free body diagram of the lamp and the cord to determine all of the forces acting on the lamp while the elevator is coming to a stop.
 
  • #3
Xenoned said:
Hi,

If a statement something like this is given:
"A body is going up with a constant acceleration of 2 m/s^2"

Does it mean that acceleration due to gravity acts on it and we have to subtract 9.8 from 25?

We subtract forces by Newton's laws of motion right?

There is a question like this:
"A lamp hangs vertically from a chord in a descending lift. The lift has a deceleration of 5.2 m/s2 before coming to a halt. If the tension in cord is 30 N, find the mass of the lamp."

We answer it by T=m(g+a) ; a= acceleration/deceleration and get the answer.
but here we use g+a .

why not just acceleration?
Why shouldn't we subtract or add g in case of 1D motion?
Those are two contradictory questions! You ask why we use g at all in the first question then ask why we shouldn't use it in the second question!
 
  • #4
I am sorry. It's not 25 but 2.
 
  • #5
Xenoned said:
Hi,

If a statement something like this is given:
"A body is going up with a constant acceleration of 2 m/s^2"

Does it mean that acceleration due to gravity acts on it and we have to subtract 9.8 from 25?

The force of gravity acts on it. The net force on the body produces the given acceleration of 2 m/s2 upwards. The net force is comprised of the gravitational force and the force exerted by whatever mechanism is propelling the body upward.
 

FAQ: Acceleration upwards and the effect of 'g'

What is acceleration upwards?

Acceleration upwards refers to the rate at which an object's speed increases while moving in an upward direction.

What is the effect of 'g' on acceleration upwards?

'g' is the symbol for acceleration due to gravity. The effect of 'g' on acceleration upwards is that it causes objects to accelerate towards the Earth at a rate of 9.8 meters per second squared.

How is acceleration upwards calculated?

Acceleration upwards can be calculated using the formula a = F/m, where 'a' is the acceleration, 'F' is the net force acting on the object, and 'm' is the mass of the object.

What factors can affect acceleration upwards?

The factors that can affect acceleration upwards include the net force acting on the object, the mass of the object, and any external forces such as air resistance or friction.

What are some real-life examples of acceleration upwards?

Examples of acceleration upwards in real-life include objects being thrown into the air, a rocket taking off from the ground, and a person jumping off a diving board into a pool.

Back
Top