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Anastaziya [24]
3 years ago
14

An 87.6 g lead ball is dropped from rest from a height of 7.00 m. The collision between the ball and the ground is totally inela

stic. Assuming all the ball's kinetic energy goes into heating the ball, find its change in temperature.
Physics
1 answer:
bija089 [108]3 years ago
7 0

Taking specific heat of lead as 0.128 J/gK = c

We have energy of ball at 7.00 meter height = mgh = 87.6*10^{-3}*9.81*7

When leads gets heated by a temperature ΔT energy needed = mcΔT

                                                                      = 87.6*10^{-3}*0.128*10^3ΔT

Comparing both the equations

                      87.6*10^{-3}*9.81*7 = 87.6*10^{-3}*0.128*10^3ΔT

                        ΔT = 0.536 K

                        Change in temperature same in degree and kelvin scale

                                      So ΔT = 0.536 ^0C

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Answer:

Stretch can be obtained using the Elastic potential energy formula.

The expression to find the stretch (x) is x=\sqrt{\frac{2\times EPE}{k}}

Explanation:

Given:

Elastic potential energy (EPE) of the spring mass system and the spring constant (k) are given.

To find: Elongation in the spring (x).

We can find the elongation or stretch of the spring using the formula for Elastic Potential Energy (EPE).

The formula to find EPE is given as:

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

Rewriting the above expression in terms of 'x', we get:

x=\sqrt{\frac{2\times EPE}{k}}

Example:

If EPE = 100 J and spring constant, k = 2 N/m.

Elongation or stretch is given as:

x=\sqrt{\frac{2\times EPE}{k}}\\\\x=\sqrt{\frac{2\times 100}{2}}\\\\x=\sqrt{100}=10\ m

Therefore, the stretch in the spring is 10 m.

So, stretch in the spring can be calculated using the formula for Elastic Potential Energy.

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Block with mass m =7.6 kg is hung from a vertical spring. when the mass hangs in equilibrium, the spring stretches x = 0.29 m. w
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1) 256.9 N/m

The force applied to the spring is equal to the weight of the block hanging on the spring:

F=mg=(7.6 kg)(9.8 m/s^2)=74.5 N

And the spring constant can be found by using Hook's law, because we know that the displacement caused by this force is x = 0.29 m:

F=kx\\k=\frac{F}{x}=\frac{74.5 N}{0.29 m}=256.9 N/m

2) 1.08 Hz

The angular frequency of oscillation of the spring is given by the formula:

\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{256.9 N/m}{7.6 kg}}=5.81 rad/s

And the frequency of oscillation is given by:

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3) 2.19 m/s

The velocity at time t of the block is given by:

v=v_0 cos (\omega t)

where

v_0 = 4.4 m/s is the initial velocity of the block

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t is the time

Substituting t=0.36 s, we find the speed of the block at that time:

v(0.36 s)=(4.4 m/s) ( cos ((5.81 rad/s)(0.36 s)) = -2.18 m/s

And the negative sign means that the direction of the velocity is upward (because the initial velocity was downward)

4) 25.6 m/s^2

The maximum acceleration is given by:

a_0 = \omega^2 A

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We can find the amplitude by using the law of conservation of energy: in fact, the kinetic energy at the equilibrium point must be equal to the elastic potential energy at the point of maximum displacement:

K=U\\\frac{1}{2}mv_0^2 = \frac{1}{2}kA^2\\A=\sqrt{\frac{mv_0^2}{k}}=\sqrt{\frac{(7.6 kg)(4.4 m/s)^2}{256.9 N/m}}=0.76 m

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5) 95.1 N

The magnitude of the net force acting on the block is given by the difference between the weight and the restoring force of the spring:

F=mg-kx

First, we need to find the position x at t=0.36 s, which is given by

x(t)=A sin(\omega t)=(0.76 m)(sin ((5.81 rad/s)(0.36 s))=0.66 m

And so, the net force is

F=(7.6 kg)(9.8 m/s^2)-(256.9 N/m)(0.66 m)=-95.1 N

And the negative sign means the direction of the force is upward.

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3 years ago
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