Finding acceleration and weight

[lim]

Homework Statement

On some strange planet you find that a 9.2 kg object falls downward 27.4 meters in 6.2 seconds. What is the magnitude of the accleration due to gravity on this planet? How much does the object weigh on this planet? How much would it weigh back on earth?

Homework Equations

xf = xi +vit +1/2at^2
w= mg

The Attempt at a Solution

a. 27.4 m/ 6.2 s = 4.42 m/s

27.4 = 0 + 4.42 + 1/2 a 6.2^2

27.4 = 4.42 + 19.22a

a= 1.195 m/s^2

b. 1.195(9.2)

10.99 mg(?)

c. (1.195) (9.8 m/s)

= 11.62 mg(?)

I thought the units would be in mg for the weight, but it said my units were inapproriate as well.

 

[Astronuc]

Acceleration has units of $$\frac{distance}{time^2}$$

At sea level on earth, the gravitational acceleration is about 9.81 m/s².

a. 27.4 m/ 6.2 s = 4.42 m/s

would give the average velocity, not initial velocity.

Try y = 1/2 a t²

http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html#ffall

 

[m2003]

Good job on finding the acceleration due to gravity on this strange planet! Your calculations are correct. However, the units for weight should be in Newtons (N) instead of milligrams (mg). This is because weight is a measure of force, and the standard unit for force is Newtons. To convert from kilograms to Newtons, you can use the formula F = mg, where m is the mass in kilograms and g is the acceleration due to gravity in meters per second squared (m/s^2).

So, to find the weight of the object on this planet, we can use the formula F = (9.2 kg)(1.195 m/s^2) = 11.00 N. This means that the object weighs 11.00 N on this planet.

To find the weight of the object back on Earth, we can use the same formula, but with the acceleration due to gravity on Earth (9.8 m/s^2). So, the weight of the object on Earth would be (9.2 kg)(9.8 m/s^2) = 90.16 N. This means that the object would weigh 90.16 N back on Earth, which is significantly more than it weighs on the strange planet due to the difference in the acceleration due to gravity.

Overall, great job on your calculations and just remember to use the appropriate units for weight (Newtons) instead of milligrams. Keep up the good work!