Is there supposed to be an image if so there is none
I believe the answer is potential energy if i remember correctly.
Answer:
See the answers below.
Explanation:
to solve this problem we must make a free body diagram, with the forces acting on the metal rod.
i)
The center of gravity of the rod is concentrated in half the distance, that is, from the end of the bar to the center there is 40 [cm]. This can be seen in the attached free body diagram.
We have only two equilibrium equations, a summation of forces on the Y-axis equal to zero, and a summation of moments on any point equal to zero.
For the summation of forces we will take the forces upwards as positive and the negative forces downwards.
ΣF = 0

Now we perform a sum of moments equal to zero around the point of attachment of the string with the metal bar. Let's take as a positive the moment of the force that rotates the metal bar counterclockwise.
ii) In the free body diagram we can see that the force acts at 18 [cm] of the string.
ΣM = 0
![(15*9) - (18*W) = 0\\135 = 18*W\\W = 7.5 [N]](https://tex.z-dn.net/?f=%2815%2A9%29%20-%20%2818%2AW%29%20%3D%200%5C%5C135%20%3D%2018%2AW%5C%5CW%20%3D%207.5%20%5BN%5D)
Answer:
The velocity of the student has after throwing the book is 0.0345 m/s.
Explanation:
Given that,
Mass of book =1.25 kg
Combined mass = 112 kg
Velocity of book = 3.61 m/s
Angle = 31°
We need to calculate the magnitude of the velocity of the student has after throwing the book
Using conservation of momentum along horizontal direction


Put the value into the formula


Hence, The velocity of the student has after throwing the book is 0.0345 m/s.