work & Energy?

Question

A(n) 88 g arrow is fired from a bow whose string exerts an average force of 105 N on the arrow over a distance of 72 cm. What is the speed of the arrow as it leaves the bow?

Answer 1

m = 88g = 0.088kg; F = 105 N, d = 72cm = 0.72m

Method 1

Fd = mv^2/2

v = sqrt(2Fd/m) = sqrt(2x105x0.72/0.088) = 41.45 m/sec

Method 2

a = F/m = 1193.1818 m/sec2

The time the force is acting on the arrow is given by:

t = sqrt(2d/a) = 0.03474 sec

The velocity as it leaves the bow is given by:

V = at = 41.45 m/sec

Answer 2

The Work accomplished by the force is equal to

W = Favg * Dx = 105 * 0.72 = 75.6 J

The arrow gains this as kinetic energy, so

Ke = 1/2 m v^2 = 75.6 J

Solve this as follows: multiply by two, divide by the mass and take the square root,

v = sqrt(2 * 75.6 / 0.088) = 41.45 m/s

Answer 3

mass of the arrow=0.088 kg Force=105 N and distance=0.72m
the work is done on the arrow because of the kinetic energy
work done=force*distance=105*(0.72)
if the velocity is 'v' m/sec,then kinetic energy=1/2(0.088)v^2
equating 105(0.72)=1/2(0.088)v^2 so v^2=[2(105)(0.72)/0.088]
=>1718 and o v=(1718)^1/2=41.45 m/sec. approx.

Answer 4

Work done on arrow = Kinetic energy of arrow

F*d = (1/2)mv^2

105 [N] * .72 [m] = (1/2) * .088 [kg] * v^2

105 [ kg * m /s^2] * .72 [m] = .044 [kg] * v^2
75.6 [kg] * [m/s]^2 = .044 [kg] * v^2
v^2 = 171.8 [m/s]^2
v = 41.45 m/s
=4145 cm/s

Answer 5

Hmm, shouldn't you be doing your own homework? Not asking on here?