Question

A boxer punches a sheet of paper in midair and brings it from rest up to a speed of 21 m/s in 0.042.5. Part A What acceleration is imparted to the paper? Express your answer to two significant figures and include the appropriate units. If the mass of the paper is 0.003 kg. what force does the boxer exert on it? Express your answer to two significant figures and inndwa the appropriate units. Part C How much force does the paper exert on the boxer? Express your answer to two significant fiaurne and inar. I the appropriate units.

Answer 1

Answer: the force exerted by the boxer on the paper is approximately 1.482 N

To find the acceleration imparted to the paper, we can use the formula:

acceleration (a) = change in velocity (Δv) / time taken (Δt)

Given:

Initial velocity (u) = 0 m/s (since the paper is at rest)

Final velocity (v) = 21 m/s

Time taken (Δt) = 0.0425 s

Δv = v - u

Δv = 21 m/s - 0 m/s

Δv = 21 m/s

Using the formula for acceleration:

a = Δv / Δt

a = 21 m/s / 0.0425 s

a ≈ 494.12 m/s²

Therefore, the acceleration imparted to the paper is approximately 494.12 m/s².

Now, to find the force exerted by the boxer on the paper, we can use Newton's second law of motion:

force (F) = mass (m) * acceleration (a)

Given:

Mass of the paper (m) = 0.003 kg

Acceleration (a) = 494.12 m/s²

Using the formula for force:

F = m * a

F = 0.003 kg * 494.12 m/s²

F ≈ 1.482 N

For Part C, the paper exerts an equal but opposite force on the boxer (according to Newton's third law of motion). Therefore, the force exerted by the paper on the boxer is also approximately 1.482 N.