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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Answer:
(a) Magnitude of static friction force is 109 N
(b) Minimum possible value of static friction is 0.356
Solution:
As per the question;
Horizontal force exerted by the girl, F = 109 N
Mass of the crate, m = 31.2 kg
Now,
(a) To calculate the magnitude of static friction force:
Since, the crate is at rest, the forces on the crate are balanced and thus the horizontal force is equal to the frictional force, f:
F = f = 109 N
(b) The maximum possible force of friction between the floor and the crate is given by:
where
N = Normal reaction = mg
Thus
For the crate to remain at rest, The force exerted on the crate must be less than or equal to the maximum force of friction.