How Does a 0.0200-kg Rock Accelerate When Shot Upward with 98.0 N?

  • Thread starter Thread starter andrewy77
  • Start date Start date
  • Tags Tags
    Acceleration Rock
AI Thread Summary
A 0.0200-kg rock is shot upward with an average force of 98.0 N, and the discussion revolves around calculating its acceleration, velocity, and other dynamics while in the slingshot and after leaving it. The key formula discussed is F = ma, which allows for the determination of acceleration by dividing the net force by mass. Clarification is sought on whether the 98.0 N force is the net force or just the applied force, as this affects subsequent calculations. It is suggested that if the 98.0 N is the applied force, the gravitational force must be subtracted to find the actual acceleration. The conversation emphasizes the importance of understanding the forces acting on the rock to solve the problem accurately.
andrewy77
Messages
6
Reaction score
0

Homework Statement



A 0.0200-kg rock is shot directly upward with an average force of 98.0 n. The rock accelerates through a distance of 0.150 m.

Homework Equations



a.) Determine the acceleration while in the slingshot.
b.) What is the velocity with which it leaves the slingshot?
c.) What force acts on the rock after it leaves the slingshot and what is its value?
d.) What is the value of the acceleration/deceleration after it leaves the slingshot?
e.) How far upward does it travel before it begins to fall?
f.) How long is the rock in the air upon returning to the level of the slingshot?


The Attempt at a Solution



I've tried v= v0+ at for part a; however, my Physics teacher said I was so wrong. Could you help with with this problem? I've tried everything, and I can get the rest of the worksheet. It's just this problem that makes my head spin around like Linda Blair...

Thanks so much and have a great day! :)

Andrew
 
Physics news on Phys.org
Hi Andrewy77. Welcome to Physics Forums.

What forces are acting on the mass? What's the total force? What's the formula for acceleration given total force and mass?
 
Thanks for the greeting!

The problem doesn't say; therefore, I think the only force acting on it is gravity, which is 9.8 m/s.

Well, given that m= f/a

a= f/m. That is all the information he gave us.
 
andrewy77 said:

Homework Statement



A 0.0200-kg rock is shot directly upward with an average force of 98.0 n. The rock accelerates through a distance of 0.150 m.

Homework Equations



a.) Determine the acceleration while in the slingshot.
b.) What is the velocity with which it leaves the slingshot?
c.) What force acts on the rock after it leaves the slingshot and what is its value?
d.) What is the value of the acceleration/deceleration after it leaves the slingshot?
e.) How far upward does it travel before it begins to fall?
f.) How long is the rock in the air upon returning to the level of the slingshot?


The Attempt at a Solution



I've tried v= v0+ at for part a; however, my Physics teacher said I was so wrong. Could you help with with this problem? I've tried everything, and I can get the rest of the worksheet. It's just this problem that makes my head spin around like Linda Blair...

Thanks so much and have a great day! :)

Andrew

Newton's Laws; specifically the second.
F = ma is what you are after - you have the force, you have the mass, you can find the [average] acceleration.
 
Did you draw a Free Body Diagram for the rock while it's in the slingshot?
 
Yes; the diagram helped a little.

For part a), I used f=ma

Solving for a gets you a=f/m

With the given data, I got 4900 m/s for a. That sounds a little quirky..
 
andrewy77 said:
Yes; the diagram helped a little.

For part a), I used f=ma

Solving for a gets you a=f/m

With the given data, I got 4900 m/s for a. That sounds a little quirky..

Looks like a plausible acceleration for such a small mass. However, I have one question that should be decided before you proceed: Is the given 98.0N force the net force on the stone including the effects of gravity (weight), or is it the average force applied by the slingshot alone? If its the latter case, then the actual acceleration will be reduced by g.
 
Well, the problem states the rock was shot with 98.0 n of force, so I'm guessing it's the force applied to shoot the rock.
 
andrewy77 said:
Well, the problem states the rock was shot with 98.0 n of force, so I'm guessing it's the force applied to shoot the rock.

If that is the case then subtract g from the acceleration you calculated above:
a = \frac{f_{net}}{m} = \frac{F - m g}{m} = \frac{F}{m} - g
 
Back
Top