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3 years ago

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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?

1 answer:

stich3 [128]3 years ago

8 0

(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/ $s^{2}$

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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An object is pushed from rest across a sheet of ice, accelerating at 8.0 m/s^2 [E] over a displacement of 1.05 m [E]. The object

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Answer:

$D_T=18.567m$

Explanation:

From the question we are told that:

Acceleration $a=8.0 m/s^2$

Displacement $d=1.05 m$

Initial time $t_1=6.0s$

Final Time $t_2=2.5s$

Generally the equation for Velocity of 1.05 travel is mathematically given by

Using Newton's Law of Motion

$V^2=2as$

$V=\sqrt{2*6*1.05}$

$V=4.1m/s$

Generally the equation for Distance traveled before stop is mathematically given by

$d_2=v*t_1$

$d_2=3.098*4$

$d_2=12.392$

Generally the equation for Distance to stop is mathematically given by

Since For this Final section

Final velocity $v_3=0 m/s$

Initial velocity $u_3=4.1 m/s$

Therefore

Using Newton's Law of Motion

$-a_3=(4.1)/(2.5)$

$-a_3=1.64m/s^2$

Giving

$v_3^2=u^2-2ad_3$

Therefore

$d_3=\frac{u_3^2}{2ad_3}$

$d_3=\frac{4.1^2}{2*1.64}$

$d_3=5.125m$

Generally the Total Distance Traveled is mathematically given by

$D_T=d_1+d_2+d_3$

$D_T=5.125m+12.392+1.05 m$

$D_T=18.567m$

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