Answer:
I'm pretty sure the answer is 0 m/s²
Explanation:
The horizontal velocity of the second rock is 5 m/s, so if we pretend air resistance doesn't exist, it will maintain that horizontal velocity, meaning that there is no horizontal acceleration.
Static frictional force = ƒs = (Cs) • (Fɴ)
2.26 = (Cs) • m • g
2.26 = (Cs) • (1.85) • (9.8)
Cs = 0.125
kinetic frictional force = ƒκ = (Cκ) • (Fɴ)
1.49 = (Cκ) • m • g
1.49 = (Cκ) • (1.85) • (9.8)
Cκ = 0.0822
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.