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icang [17]
3 years ago
10

A 2.5 kg block is launched along the ground by a spring with a spring constant of 56 N/m. The spring is initially compressed 0.7

5 m.
Disregarding friction, how fast will the block move after the spring is released all the way and the block slides away from it?

3.5 m/s
4.1 m/s
13 m/s
16 m/s
Physics
2 answers:
Norma-Jean [14]3 years ago
7 0

vf ^2 = kx^2/m = 56(0.75)^2 / 2.5 = 12.6


Therefore, v= 3.5 m/s.

olga55 [171]3 years ago
4 0

Answer: The correct option is 3.5m/s.

Explanation: In the question, conservation of energy is taking place which means that the energy is getting transferred from one form to another form.

Here, elastic potential energy of the spring is getting converted to the kinetic energy of the block.

Mathematically,

\frac{1}{2}kx^2=\frac{1}{2}mv^2

where, k = spring constant = 56N/m

x = compression of extension of spring = 0.75m

m = Mass of the block = 2.5kg

v = velocity of the block = ? m/s

Putting values in above equation, we get:

\frac{1}{2}\times 56\times (0.75)^2=\frac{1}{2}\times 2.5\times v^2\\\\v=3.54m/s

Hence, the block will move at a speed of 3.5m/s

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A cannon is fired horizontally at 243 m/s off of a 62 meter tall, shear vertical cliff. How far in meters from the base of the c
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Answer:

865.08 m

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 243 m/s

Height (h) of the cliff = 62 m

Horizontal distance (s) =?

Next, we shall determine the time taken for the cannon to get to the ground. This can be obtained as follow:

Height (h) of the cliff = 62 m

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =?

h = ½gt²

62 = ½ × 9.8 × t²

62 = 4.9 × t²

Divide both side by 4.9

t² = 62/4.9

Take the square root of both side.

t = √(62/4.9)

t = 3.56 s

Finally, we shall determine the horizontal distance travelled by the cannon ball as shown below:

Initial velocity (u) = 243 m/s

Time (t) = 3.56 s

Horizontal distance (s) =?

s = ut

s = 243 × 3.56 s

s = 865.08 m

Thus, the cannon ball will impact the ground 865.08 m from the base of the cliff.

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