A student at another university repeats the experiment you did in lab. Her target ball is 0.860 m above the floor when it is in the target holder and the steel ball she uses has a mass of 0.0120 kg. She finds that the target ball travels a distance of 1.40 m after it is struck. Assume g = 9.80 m/s2. What is the kinetic energy (in joules) of the target ball just after it is struck?

11.8 Joules

Explanation:

Given:-

- The height of the target ball, si = 0.860 m

- The mass of target and steel ball, m = 0.012 kg

- The target ball travels a distance ( x ) after being struck = 1.40 m

Find:-

What is the kinetic energy (in joules) of the target ball just after it is struck?

Solution:-

- We are given the initial distance of the target ball as 0.86 m above the floor which travels a distance ( x ) after being struck.

- We will employ the one dimensional kinematic equation of motion to determine the initial velocity ( vi ) of the target ball as follows:

Where,

                  vf: The final velocity of target ball at maximum height = 0

                  g: The gravitational acceleration constant = 9.8 m/s^2

- Plug in the required parameters and evaluate the ( vi ) as follows:

- The kinetic energy ( Ek ) of an object with mass ( m ) and initial velocity ( vi ) is expressed as:

The kinetic energy of the target ball just after it is struck is 11.8 Joules.

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards