A 8.0×10−2 kg arrow hits the target at 25 m/s and penetrates 3.8 cm before stopping.
What average force did the target exert on the arrow?
What average force did the arrow exert on the target?
a.
average force did the target exert on the arrow
Work done = force * distance = energy lost
Initial kinetic energy = (1/2)mv²
work done = F*d
Equating energy lost to work done
F*d = (1/2)mv²
F=d*mv*sup2;/2
b) What do you know about Newton's third law?
.95N
To find the average force exerted by the target on the arrow, we can use the principle of impulse-momentum. The impulse experienced by an object is equal to the change in its momentum. The impulse can be calculated using the formula:
Impulse = mass × change in velocity
Given:
Mass of the arrow (m) = 8.0 × 10^(-2) kg
Change in velocity of the arrow (Δv) = final velocity (v) - initial velocity (u) = 0 m/s - (-25 m/s) = 25 m/s
Impulse = m × Δv
Impulse = 8.0 × 10^(-2) kg × 25 m/s
Now, since impulse is equal to the average force multiplied by the time taken for the force to act, we can rearrange the formula to find the average force.
Average force = Impulse / time
To calculate the time taken, we need to know the distance traveled by the arrow in order to come to a stop. The distance traveled can be determined if we know the deceleration of the arrow.
To find the deceleration, we can use the following formula:
v^2 = u^2 + 2as
Rearranging the formula, we get:
a = (v^2 - u^2) / (2s)
Given:
s = 3.8 cm (which is 0.038 m)
u = -25 m/s (negative sign indicates opposite direction of motion)
v = 0 (arrow comes to a stop)
Calculating the deceleration:
a = (0^2 - (-25)^2) / (2 × 0.038 m)
Once we have the deceleration, we can find the time taken to stop using the formula:
Change in velocity = acceleration × time
Now, with the time taken, we can calculate the average force exerted by the target on the arrow using the formula:
Average force = Impulse / time
To find the average force exerted by the arrow on the target, we can use Newton's third law which states that every action has an equal and opposite reaction. This means that the average force exerted by the arrow on the target will be equal in magnitude but opposite in direction to the average force exerted by the target on the arrow. Therefore, the magnitude of the average force exerted by the arrow on the target will be the same as the magnitude of the average force exerted by the target on the arrow.