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3 years ago

A spring extends by .16 meters when pulled with 20 N of Force. What is the Potential Energy in the spring at this extension?

Physics

1 answer:

Yuri [45]3 years ago

4 0

Answer:

Potential energy = 1.6 J

Work done = -1.6 J

Explanation:

Potential Energy in the spring at this extension?

The potential energy U in the spring is U = 1/2kx² where k = spring constant = F/x where F = force = 20 N and x = extension = 0.16 m. So k = 20 N/0.16 m = 125 N/m

U = 1/2kx²

= 1/2 × 125 N/m × (0.16 m)²

= 1.6 J

What is the work done to extend the spring. 16 meters

Work done W = U₁ - U₂ where U₁ = initial potential energy = 0 and U₂ = final potential energy = 1.6 J

W = 0 - 1.6 J

= -1.6 J

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a 2.0 kg block slides on the horizontal, frictionless surface until it counters a spring force constant with

icang [17]

Complete question:

a 2.0 kg block slides on the horizontal, frictionless surface until it counters a spring with force constant of 955 N/m. The block comes to rest after compressing the spring a distance of 4.6 cm. Find the initial speed (in m/s) of the block.

Answer:

The initial speed of the block is 1.422 m/s

Explanation:

Given;

mass of the block, m = 2.0 kg

force constant of the spring, K = 955 N/m

compression of the spring, x = 4.6 cm = 0.046 m

Apply Hook's law to determine applied force on the spring;

F = Kx

F = (955 N/m)(0.046 m)

F = 43.93 N

Apply Newton's 2nd law to determine the magnitude of deceleration of the block when it encounters the spring;

F = ma

a = F / m

a = 43.93 / 2

a = 21.965 m/s²

Apply kinematic equation to determine the initial speed of the block;

v² = u² + 2ax

where;

v is the final speed of the block = 0

u is the initial speed of the block

x is the distance traveled by the block = compression of the spring

a is the block deceleration = -21.965 m/s²

0 = u² + 2(-21.965 )(0.046)

0 = u² - 2.021

u² = 2.021

u = √2.021

u = 1.422 m/s

Therefore, the initial speed of the block is 1.422 m/s

8 0

3 years ago

A thin block of soft wood with a mass of 0.078 kg rests on a horizontal frictionless surface. A bullet with a mass of 4.67 g is

natali 33 [55]

Answer:

Final speed of the bullet is 228.3 m/s

Explanation:

As we know that there is no external force on the system of wooden block and the bullet

so we can say momentum of the system is conserved here

so here we can say

$P_i = P_f$

$m_1v_1 = m_1v_{1f} + m_2v_{2f}$

$4.67\times 10^{-3}(613) = 4.67 \times 10^{-3}v_{1f} + (0.078)(23)$

so we will have

$2.86 = 4.67\times 10^{-3} v_{1f} + 1.794$

$v_{1f} = 228.3 m/s$

7 0

3 years ago

The boy on the tower throws a ball 20m downrange. What is his pitching speed?

san4es73 [151]

Answer:

The pitching speed of the ball is 19.7 m/s

Explanation:

Here, we can use the third equation of motion, $v^{2} = u^{2} - 2as$

whereas v represents the final velocity, u represents initial velocity, a is the acceleration due to gravity and s is the displacement or distance an object traveled

Here, the initial velocity of the the ball is given as zero and the acceleration due to gravity is 9.8 , the distance 's' is given as 20 m

Using the equation, $v^{2} = 2 * 9.8 * 20 = 392\\v = \sqrt{392} = 19.7m/s$

Hence, the pitching speed of the ball is 19.7 m/s

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3 years ago

Select the correct answer.

Stels [109]

Answer:

An electric bell is placed inside a transparent glass jar. The bell can be turned on and off using a switch on the outside of the jar. A vacuum is created inside the jar by sucking out the air. Then the bell is rung using the switch. What will we see and hear?

We’ll see the bell move, but we won’t hear it ring.

We won’t see the bell move, but we’ll hear it ring.

We’ll see the bell move and hear it ring.

We won’t see the bell move or hear it ring.

We’ll see the sound waves exit the vacuum pump.

Explanation:

so, the answer to the question is

We'll see the bell move, but we won’t hear it ring.

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3 years ago

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lorasvet [3.4K]

Answer:

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