Acceleration of a Rock Problem

In summary, the problem involves a 0.0200-kg rock being shot upward with an average force of 98.0 N. The rock accelerates through a distance of 0.150 m. The task is to determine the acceleration while in the slingshot, the velocity with which it leaves the slingshot, the force acting on the rock after it leaves the slingshot and its value, the value of the acceleration/deceleration after it leaves the slingshot, the distance traveled upward before it begins to fall, and the time the rock is in the air upon returning to the level of the slingshot. To solve this problem, one must consider Newton's Laws, specifically the second law, F
  • #1
andrewy77
6
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
 
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  • #2
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?
 
  • #3
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.
 
  • #4
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.
 
  • #5
Did you draw a Free Body Diagram for the rock while it's in the slingshot?
 
  • #6
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..
 
  • #7
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.
 
  • #8
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.
 
  • #9
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:
[tex] a = \frac{f_{net}}{m} = \frac{F - m g}{m} = \frac{F}{m} - g[/tex]
 

1. What is the definition of acceleration?

Acceleration is the rate at which an object's velocity changes. It is a vector quantity, meaning it has both magnitude (how much the velocity changes) and direction (the direction in which the velocity changes).

2. How is acceleration calculated?

Acceleration can be calculated by dividing the change in velocity by the change in time. The equation for acceleration is a = (vf - vi) / t, where a is acceleration, vf is final velocity, vi is initial velocity, and t is time.

3. How does acceleration of a rock differ from acceleration of other objects?

The acceleration of a rock is affected by the force of gravity acting on it. On Earth, the acceleration due to gravity is approximately 9.8 m/s^2. This means that a rock dropped from a height will experience an acceleration of 9.8 m/s^2 towards the ground.

4. Can acceleration be negative?

Yes, acceleration can be negative. This means that an object is slowing down or its velocity is decreasing. For example, if a car is driving east at 50 m/s and then turns around and starts driving west at 30 m/s, its acceleration would be negative because its velocity is decreasing.

5. How does mass affect acceleration?

According to Newton's Second Law of Motion, the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. This means that a larger mass will require more force to accelerate at the same rate as a smaller mass.

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