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3 years ago

Why would the time of flight depend on the angle of the launch

1 answer:

3 0

The time of flight depends on the vertical speed. The angle of the launch determines how much of the total launch velocity is horizontal speed and how much of it is vertical speed.

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Diesel fuel is used in the engine of trucks that carry dirt fruit and other cargo. Fuel is burned in engines to make the motor m

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3 years ago

A horizontal block-spring system with the block on a frictionless surface has total mechanical energy E = 50.9 J and a maximum d

Llana [10]

(a) 2446 N/m

When the spring is at its maximum displacement, the elastic potential energy of the system is equal to the total mechanical energy:

$E=U=\frac{1}{2}kA^2$

where

U is the elastic potential energy

k is the spring constant

A is the maximum displacement (the amplitude)

Here we have

U = E = 50.9 J

A = 0.204 m

Substituting and solving the formula for k,

$k=\frac{2E}{A^2}=\frac{2(50.9)}{(0.204)^2}=2446 N/m$

(b) 50.9 J

The total mechanical energy of the system at any time during the motion is given by:

E = K + U

where

K is the kinetic energy

U is the elastic potential energy

We know that the total mechanical energy is constant: E = 50.9 J

We also know that at the equilibrium point, the elastic potential energy is zero:

$U=\frac{1}{2}kx^2=0$ because x (the displacement) is zero

Therefore the kinetic energy at the equilibrium point is simply equal to the total mechanical energy:

$K=E=50.9 J$

(c) 8.55 kg

The maximum speed of the block is v = 3.45 m/s, and it occurs when the kinetic energy is maximum, so when

K = 50.9 J (at the equilibrium position)

Kinetic energy can be written as

$K=\frac{1}{2}mv^2$

where m is the mass

Solving the equation for m, we find the mass:

$m=\frac{2K}{v^2}=\frac{2(50.9)}{(3.45)^2}=8.55 kg$

(d) 2.14 m/s

When the displacement is

x = 0.160 m

The elastic potential energy is

$U=\frac{1}{2}kx^2=\frac{1}{2}(2446)(0.160)^2=31.3 J$

So the kinetic energy is

$K=E-U=50.9 J-31.3 J=19.6 J$

And so we can find the speed through the formula of the kinetic energy:

$K=\frac{1}{2}mv^2 \rightarrow v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(19.6)}{8.55}}=2.14 m/s$

(e) 19.6 J

The elastic potential energy when the displacement is x = 0.160 m is given by

$U=\frac{1}{2}kx^2=\frac{1}{2}(2446)(0.160)^2=31.3 J$

And since the total mechanical energy E is constant:

E = 50.9 J

the kinetic energy of the block at this point is

$K=E-U=50.9 J-31.3 J=19.6 J$

(f) 31.3 J

The elastic potential energy stored in the spring at any time is

$U=\frac{1}{2}kx^2$

where

k = 2446 N/m is the spring constant

x is the displacement

Substituting

x = 0.160 m

we find the elastic potential energy:

$U=\frac{1}{2}kx^2=\frac{1}{2}(2446)(0.160)^2=31.3 J$

(g) x = 0

The postion at that instant is x = 0, since it is given that at that instant the system passes the equilibrium position, which is zero.

4 0

3 years ago