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A 25.0 kg projectile is fired by accelerating it with an electromagnetic rail gun on the earth's surface. the rail makes a 30.0 degree angle with the horizontal, and the gun applies a 1250 n force on the projectile for a distance of 7.50 m along the rail. (a) ignoring air resistance and friction, what is the net work done on the projectile, by all the forces acting on it, as it moves 7.50 m along the rail? (b) assuming it started at rest, what is its speed after it has moved the 7.50 m?

Answer :

(a)

The work done on the projectile is 9375 joule.

The work on the projectile is calculated as

W=F×d

=1250×7.5

=9375 joule

(b)

The speed of the projectile after 7.5 m is 27.38 m/s

First we need to find out the acceleration of the projectile

F=m×a

1250=25×a

a=50 m/[tex] s^{2} [/tex]

Now the velocity of the projectile after 7.5 m is calculated as

v^2=u^2+2a×s

v^2=0+2×50*7.5

v=27.38 m/s

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