4. A 200 g mass is attached to a spring of spring constant k. The spring is comp

Question

4. A 200 g mass is attached to a spring of spring constant k. The spring is compressed 15 cm from its equilibrium value. When released the mass reaches a speed of 5 m/s. What is the spring constant (in N/m)?


5. A 34-g bullet traveling at 120m/s embeds itself in a wooden block on a smooth surface. The block then slides toward a spring and collides with it. The block compresses the spring (k=100 N/m) a maximum of 1.25 cm. Calculate the mass of the block of wood.

10. A net torque of 36N*m acts on a wheel rotating about a fixed axis for 6 s. During this time the angular speed of the wheel increases from 0 to 12 rad/s. The applied force is then removed, and the wheel comes to rest in 75 s.
a. What is the moment of inertia of the wheel?
b. What is the magnitude of the frictional torque?
c. How many revolutions does the wheel make?

Explanation / Answer

4 .

1/2 m v^2 = 1/2 kx^2

m v^2 = kx^2 ;

0.2 * 5 ^2 = k * 0.15^2 ;

k = 222.22 N / m <---------ans

5 .

conservation of energy ,

1/2 ( M + m ) V^2 = 1/2 k x^2 ;

( M + m ) V^2 = k x^2 ;

( M + 0.034 ) V^2 = 100 * ( 1.25e-2 )^2 ;

( M + 0.034 ) V^2 = 0.015625 ----------------------1

conservation of momentum .

m v = ( M + m ) V ;

0.034 * 120 = ( M + 0.034 ) V ;

( M + 0.034 ) V = 4.08 ---------------------------2

dividin 1 by 2 ,

V = 0.015625 / 4.08 = 3.829 e -3 m /s

putting this in 2 ,

M = 1065.3356 kg <----------ans

10.

= 0 + t ;

12 = 0 +   * 6 ;

or   = 2 rad / s^2 ;

= I *

36 = I * 2 ;

( a )

I = 18 ; <------------ans

( b )

= 0 + t ;

0 = 12 + * 75

= -0.16 ;

F rictional torque = I   = 18 * -0.16

frictional torque = -2.88 N m <------------ans

( c )

for the accelerating part ,

= 0 t + 1/2 t^2 ;

= 0 * t + 0.5 * 2 * 6^2 =36 rad s ;

for the deccelerating part ,

= 12 * 75 - 0.5 * 0.16 * 75^2 = 450 rad ;

total rad = 36 + 450 = 486 rad .

revolutions= 486 / [ 2 ] = 77.349 revolutions <---------ans

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