a boxer punches a sheet of paper in mid air, bringing it from rest up to a speed of 25 m/s in 0.05 second. The mass of the paper is 0.003 kg. Show that the force of the punch on the paper is only 1.5 N.
Force = rate of change of momentum
= mass * (change in velocity) / time
Can you take it from here?
still need help
.25 N
To determine the force of the punch on the paper, we can use Newton's second law of motion, which states that force (F) is equal to the mass (m) of an object multiplied by its acceleration (a). Mathematically, this can be expressed as:
F = m * a
Given:
Mass of the paper (m) = 0.003 kg
Acceleration (a) = Change in velocity (Δv) / Time taken (Δt)
We are provided with the velocity change (from rest to 25 m/s) and the time it takes to reach that velocity.
Δv = Final velocity (v_f) - Initial velocity (v_i) = 25 m/s - 0 m/s = 25 m/s
Δt = 0.05 s
Using these values, we can calculate the acceleration:
a = Δv / Δt
a = 25 m/s / 0.05 s
a = 500 m/s²
Now, we can substitute the mass and acceleration into the formula for force:
F = m * a
F = 0.003 kg * 500 m/s²
F = 1.5 N
Hence, the force of the punch on the paper is 1.5 N.