Question

Please help, physics question:

A 0.155 kg arrow is shot upward at 31.4 m/s. What is its kinetic energy (KE) when it is 30.0 m above the ground?

thank you for your help in advance!​

Answer 1

To find the kinetic energy of an arrow when it is 30.0 m above the ground, we use the conservation of energy principle by subtracting the gravitational potential energy it has gained from its initial kinetic energy.

Step-by-step explanation:

To calculate the kinetic energy (KE) of a 0.155 kg arrow shot upward at 31.4 m/s when it is 30.0 m above the ground, we need to consider both the kinetic energy of the arrow and the potential energy that the arrow has gained due to its height above the ground.

The total mechanical energy (assuming no air resistance) is the sum of kinetic and potential energy and remains constant, i.e.,

Total mechanical energy = KE + PE

At the time of launch, all of the arrow's mechanical energy is kinetic since it is at ground level (we are neglecting any differences in height from which the arrow was shot).

The initial kinetic energy (KE_initial) of the arrow can be calculated using:

KE_initial = (1/2) * mass * velocity^2

After the arrow has risen to 30.0 m above the ground, it will have gained gravitational potential energy (PE) given by:

PE = mass * gravitational acceleration (g) * height

Since energy is conserved, the kinetic energy at 30.0 m (KE_final) can be found by:

KE_final = Total mechanical energy - PE

We calculate KE_final by plugging in the values of mass, height, and using 9.8 m/s^2 for g:

PE = 0.155 kg * 9.8 m/s^2 * 30.0 m

And subtracting this PE from the initial KE to get KE_final:

Calculate initial KE with the given velocity and mass.

Calculate PE at 30.0 m height.

Subtract the PE from the initial KE for KE_final (the kinetic energy at 30.0 m).

Answer 2

Kinetic energy (K.E) when it is 30.0 m above the ground is 30.84 J .

Step-by-step explanation:

Given :

Mass of arrow , m = 0.155 kg .

Initial velocity , u = 31.4 m/s .

To Find :

Kinetic energy (K.E) when it is 30.0 m above the ground .

Solution :

Let , us assume upward direction be positive and downward direction is negative .

Now , by equation of motion :

`v^2-u^2=2as`

Here ,
`a=g=-9.8\ m/s^2` ( Because acceleration due to gravity is in downward direction )

Putting all given values in above equation , we get :

`v^2-31.4^2=2* -9.8* 30\\\\v^2=31.4^2-588\\\\v^2=397.96\\\\v=19.95\ m/s`

Now , kinetic energy is given by :

`K.E=(1)/(2)mv^2`

Putting all values we get :

K.E = 30.84 J .