Question

A dart gun has a spring with a constant of 875 N/m. The spring is initially compressed 5.5 cm. You fire a 0.075-kg dart from a height of 0 m. The dart hits a wall at a height of 1.5 m. Ignore air resistance.

What is the initial potential energy of the spring?

What is the speed of the dart just before it hits the wall?

The dart then penetrates 0.5 cm into the wall before coming to rest.

How much work does the wall do on the dart?

What is the magnitude of the force that the wall exerts on the dart while bringing it to rest?

Answer 1

inital potential energy of spring U = (1/2)*k*x^2

Ui = (1/2)*875*0.055^2 = 1.323 J

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initial kinetic energy Ki = 0

total energy before hitting the wall Ef = (1/2)*m*v^2 + m*g*h

from conservation of energy

Ef = Ei

(1/2)*m*v^2 + m*g*h = 1.323

((1/2)*0.075*v^2) + (0.075*9.8*1.5) = 1.323

speed v = 2.42 m/s

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work done = change in KE

W = -(1/2)*m*v^2 = -0.22 J

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work = -F*x

-0.22 = -F*0.005

F = 44 N <<<<<<<<=============ANSWER