Physics, 26.03.2020 19:34 ems71

A 15 kg uniform disk of radius R = 0.25 m has a string wrapped around it, and a m = 3.4 kg weight is hanging on the string. The system of the weight and disk is released from rest. 1)When the 3.4 kg weight is moving with a speed of 1.1 m/s, what is the kinetic energy of the entire system? K. E. =

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Answer from: nagwaelbadawi

5.45 J

Explanation:

When the 3.4 kg weight is moving with a speed of 1.1 m/s, what is the kinetic energy of the entire system?

RKE = $\frac{1}{2}I \omega ^2$

where;

$I = \frac{1}{2} mr^2$

$I = \frac{1}{2}*15*0.25^2$

I = 0.46875 kg.m^2

$\omega = \frac{1.0}{0.25}$

$\omega = 4 rad/s$

RKE = $\frac{1}{2}I \omega ^2$

= $\frac{1}{2} *0.46875*4^2$

= 3.75 J

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

= $\frac{1}{2} *3.4*1.0^2$

= 1. 7 J

K.E = RKE + LKE

K. E = ( 3.75 + 1.7 ) J

K . E = 5.45 J

Answer from: TheOriginal2x

The answer is 6.6Joules

Explanation:

The kinetic energy of the entire system = the rotational kinetic energy of the disk + the kinetic energy of the body.

That is, K.E = RKE + BKE

The formula of a rotational kinetic energy is given as:

RKE = $1/2Iw^{2}$ -----------------------1Where:

RKE = Rotational Kinetic energy,

I = Inertia of the disk

w = angular velocity

for inertia the formula is given as , I = $mr^{2}$ ----------------2

where:

m = mass and r is the radius

Substituting the values into equation 2, we have

I = 1/2 $mr^{2}$

=1/2* 15 * 0.25^2

= 1/2 *15 * 0.0625

= 0.46875 $kgm^{2}$

Also for w which is angular velocity, we have

w = v/r-----------------------------------------------3

Where v is the velocity and r is the radius.

Substituting the values into equation 3, we have

w = v/r

= 1.1/0.25

= 4.4rads

Now putting the calculated values into the main equation 1, we have

RKE = $1/2Iw^{2}$ = 1/2 * 0.9375 * 4.4^2 = 1/2 * 0.46875 * 19.36 = 1/2 * 9.075 = 4.5375JCalculating the kinetic energy of the body BKE, we have:BKE = 1/2mv^2 1/2 * 3.4 * 1.1^2 = 1/2 * 3.4 * 1.21 = 1/2 * 4.114 = 2.057 JTherefore the kinetic energy of the entire system = RKE + BKE = 4.5375 + 2.057 = 6.59J = 6.6J approximately

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