the force exerted by a certain bow on an arrow decreases linearly after the arrow is released by the archer, starting with the value of F=275N when the bow is fully drawn and decreasing to F=0 as the arrow leaves the bow string. the tail of the arrow moves a distance of 0.5 m as the arrow is shot. Find the final speed of the arrow, which has a mass 3x10^-2 kg.

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Solution

we know that force is given as
F = ma
or
acceleration will be
a = F/m
here,
F = 275 N
m = 3x10^-2 kg
so,
a = 275 / 3x10^-2
thus,
a = 9166.66 m/s²
also
v² - u² =2as
here
u = 0
s = 0.5m
so,
v² = 2x0.5x9166.66 = 9166.66
thus, velocity will be
v = 95.74 m/s

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we know that force is given asF = maoracceleration will bea = F/mhere,F = 275 Nm = 3x10-2 kgso,a = 275 / 3x10-2thus,a = 9166.66 m/s2alsov2 - u2 =2ashereu = 0s = 0.5mso,v2 = 2x0.5x9166.66 = 9166.66thus, velocity will bev = 95.74 m/s

we know that force is given as
F = ma
or
acceleration will be
a = F/m
here,
F = 275 N
m = 3x10^-2 kg
so,
a = 275 / 3x10^-2
thus,
a = 9166.66 m/s²
also
v² - u² =2as
here
u = 0
s = 0.5m
so,
v² = 2x0.5x9166.66 = 9166.66
thus, velocity will be
v = 95.74 m/s