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2 years ago

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If you slide down a rope, it's possible to create enough thermal energy to burn your hands or your legs where they grip the rope

. Suppose a 40 kg child slides down a rope at a playground, descending 1.5 m at a constant speed. How much thermal energy is created as she slides down the rope? Assume that all potential energy can be transformed into thermal energy.

1 answer:

WARRIOR [948]2 years ago

3 0

Answer:

588J

Explanation:

Descending at constant speed means that the kinetic energy remained constant, so the potential energy lost was mostly transformed into thermal energy (we will assume all of it). The difference on potential energy is:

$\Delta U=U_f-U_i=mgh_f-mgh_i=mg(h_f-h_i)=(40kg)(9.8m/s^2)(-1.5m)=-588J$

So this potential energy lost is the thermal energy gained.

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A particle decelerates uniformly from a speed of 30 cm/s to rest in a time interval of 5.0 s. It then has a uniform acceleration

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Answer:

V=20cm/s

Explanation:

The average speed is the distance total divided the time total:

$V=X/T$

First stage:

T1=5s

$v_{f} =v_{o} - at$

But, $v_{f} =0$ (decelerates to rest)

then: $a =v_{o} /t=0.3/5=0.06m/s^{2}$

on the other hand:

$x =v_{o}*t - 1/2*at^{2}=0.3*5-1/2*0.06*5^{2}=0.75m$

X1=75cm

Second stage:

T2=5s

$x =v_{o}*t + 1/2*at^{2}=0+1/2*0.1*5^{2}=1.25m$

X2=125cm

Finally:

X=X1+X2=200cm

T=T1+T2=10s

V=X/T=20cm/s

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So we will first find the electrostatic potential at the center of the square

So here it is given as

$V = 4\frac{kQ}{r}$

here

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$r = \frac{a}{\sqrt2}$

$r = \frac{10nm}{\sqrt2} = 7.07 nm$

$Q = e = -1.6 * 10^{-19} C$

now the net potential is given as

$V = \frac{4 * 9*10^9 * (-1.6 * 10^{-19})}{7.07 * 10^{-9}}$

$V = 0.815 V$

now potential energy of alpha particle at this position

$U_i = qV = 2*1.6 * 10^{-19} * (-0.815) = -2.6 * 10^{-19} J$

Now at the mid point of one of the side

Electrostatic potential is given as

$V = 2\frac{kQ}{r_1} + 2\frac{kQ}{r_2}$

here we know that

$r_1 = \frac{a}{2} = 5 nm$

$r_2 = \sqrt{(a/2)^2 + a^2} = \frac{\sqrt5 a}{2}$

$r_2 = 11.2 nm$

now potential is given as

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Now work done in this process is given as

$W = U_f - U_i$

$W = (-0.267 * 10^{-19}) - (-0.26 * 10^{-19}}$

$W = -7 * 10^{-22} J$

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