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
1 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/
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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<span>http://d2vlcm61l7u1fs.cloudfront.net/media%2Fdd5%2Fdd5b98eb-b147-41c4-b2c8-ab75a78baf37%2FphpEgdSbC....
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Solution:
(a) The work done by the spring is given by
where k is the elastic constant of the spring and
(b) The work done by the weight is the product of the component of the weight parallel to the inclined plane and the displacement of the cart:
where the negative sign is given by the fact that
therefore, the work done by the weight is