Find the gravitational field strength of planet X

[ellieee]

Homework Statement

find gravitational field strength of planet X

Relevant Equations

weight/mass = gravitational field strength.

weight/mass = gravitational field strength.
my working is ->
weight = 150kgx10m/s² = 1500N
mass = 150kg
gravitational field strength= 10N/kg.
is this correct?

 

[BvU]

Not clear what the question is, but the answer is not correct.
How did you determine the weight ?

$\$

 

[Charles Link]

You need to calculate the acceleration in free fall from the graph that was provided. This will give you the $g$ in the formula $W=mg$. In any case, your answer is incorrect.

 

[ellieee]

You need to calculate the acceleration in free fall from the graph that was provided. This will give you the $g$ in the formula $W=mg$. In any case, your answer is incorrect.

do I calculate it using the part of the graph where the rocket is experiencing negative velocity ?

 

[Charles Link]

The graph is a graph of velocity vs. time. The acceleration will be negative and is the same anywhere on the straight line segment after the rocket has burnt out, even when the velocity is negative.

 

[ellieee]

hm so how do I calculate the acceleration from the graph then ?

 

[Charles Link]

Pick two points on the line segment: acceleration $a=\frac{v_2-v_1}{t_2-t_1}$.

 

[ellieee]

a

Pick two points on the line segment: acceleration $a=\frac{v_2-v_1}{t_2-t_1}$.

any 2 points ? but the acceleration is not constant throughout though?

 

[Charles Link]

On a straight line portion (of the velocity vs. time graph) the acceleration is constant. It is positive on the ascending straight line segment, where the fuel is propelling the rocket, and then it goes negative, and stays constant in the entire descending straight line segment. (By descending, I'm referring to the change in velocity. The slope of the segment of velocity vs. time is downward, even when the velocity is still positive).

 

[ellieee]

On a straight line portion (of the velocity vs. time graph) the acceleration is constant. It is positive on the ascending straight line segment, where the fuel is propelling the rocket, and then it goes negative, and stays constant in the entire descending straight line segment. (By descending, I'm referring to the change in velocity. The slope of the segment of velocity vs. time is downward, even when the velocity is still positive).

so I pick any 2 points from 20-90s?

 

[Charles Link]

Yes, go ahead and compute it. If you are skeptical that you will get the same answer, try 2 pairs of points.

Note: This is basically an exercise in algebra. You may not remember it, but the slope on a straight line is $m=\frac{y_2-y_1}{x_2-x_1}$.