A 0.075-kg arrow is fired horizontally. The bowstring exerts a force on the arrow over a distance of 0.90 m. the bow at 40 m/s. What average force does the bow apply to arrow?

The magnitude of average force applied by the bow to the arrow is 66.67 N.

Given data:

The mass of arrow is, m = 0.075 kg.

The distance covered by the arrow is,  s = 0.90 m.

The speed of bow is, v = 40 m/s.

In this problem, the work - energy theorem can be applied. As per the work -energy theorem, the work done by the applied force is equal to the kinetic energy change.

Here, F is the magnitude of average force.

Solving as,

Thus, we can conclude that the magnitude of average force applied by the bow to the arrow is 66.67 N.

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Use work-energy to solve this

The work done by on the bowstring will be converted into kinetic energy of the arrow

so W = F*d = K = 1/2m*v^2

o v = sqrt(2*F*d/m) = sqrt(2*65N*0.90m/0.075kg) = 39.5 m/s

Explanation:

answer: 802.3 kilo joules of energy represents 191.75 kilo calorie of energy.

explanation:

energy released during combustion of one mole methane = 802.3 kilo joules

if 1 kilo joule = 0.239006   kilo calorie

then ,802.3 kilo joule will be = 802.30.239006 kilo calorie

802.3 kilo joule = 191.75 kilo calorie

hence, the energy in kilo calorie is 191.75 kilo calorie.