Statistics:High: 100%

Average: 50%Low: 0

**Possibly useful information:**

s = s |
impulse = impulse = W =
W = K K = ( E = K + U W F Us = ( U |

1. An air puck of mass 0.25 kg is tied to a string and allowed to revolve in a circle of radius 0.5 m on a frictionless, horizontal table. The other end of the string passes through a hole in the center of the table and a mass of 1.2 kg is tied to it. The suspended mass remains in equilibrium while the puck on the tabletop revolves. What are

(a) the tension in the string,

(b) the central force exerted on the puck, and

(c) the speed of the puck?

From the forces on the T = (1.2 kg) (9.8 m/s T = 11.8 N This tension is also the central force exerted on the
puck; we call this the The puck That T = m v 11.8 N = (0.25 kg) v v v

**hanging mass,** we can
immediately write^{2})

**centripetal force**.**also** has the force of gravity, its weight
w_{w}, pulling **down** on it and the normal
force F_{n}, pushing **up** on it. There is no
net force in the vertical direction; these two forces just
cancel. But, in the horizontal direction, the tension T
provides the net force on it; we call this force the
**centripetal force.**
**centripetal force** on the puck is connected
with the puck's mass and its speed:_{c} = m v^{2} / r
^{2} / r^{2} / 0.5 m^{2} = (11.8 N) (0.5 m) / 0.25 kg^{2} = 23.52 m^{2}/s^{2}**v = 4.85 m/s****v = 4.85
m/s**

2. An Atwood's machine has a 3.00 kg mass and a 2.00-kg mass at ends of the string . The 2.00-kg mass is released from rest on the floor, 4.00 m below the 3.00-kg mass.

If the pulley is frictionless, what will be the speed of the masses when they pass each other? This was an early homework problem. At the time we solved it the first time, our main tool or connection between work and KE wasW_{net}= KEW

_{net}= W_{2}+ W_{3}= - (2 kg) (9.8 m/s^{2}) (2 m) + (3 kg) (9.8 m/s^{2}) (2 m)W

_{net}= 19.6 JThe small w's in the diagram -- w

_{2}and w_{3}-- are the weights;w

_{2}= m_{2}g = (2.0 kg) (9.8 m/s^{2})w

_{3}= m_{3}g = (2.0 kg) (9.8 m/s^{2})The capital W's in the equation above -- W

_{2}and W_{3}-- are the amounts of work done on the 2.0-kg mass and the 3.0-kg massby gravity.Only the force of gravity does work on "the system". We need not -- can not -- consider the tension in the cord for that is an "internal force" just like the forces that hold the blocks together.

The blocks have a common velocity v,

KE = 0.5 (2.0 kg) v

^{2}+ 0.5 (3.0 kg) v^{2}= (2.5 kg) v^{2}KE = (2.5 kg) v

^{2}= 19.6 J = W_{net}v

^{2}= 7.84 m^{2}/ s^{2}v = 2.8 m / s

We can describe this same situation now in terms of

Conservation of Energy;E _{f}= E_{i}K

_{Tot,f}+ U_{Tot,f}= K_{Tot,i}+ U_{Tot,i}K

_{1f}+ K_{2f}+ U_{1f}+ U_{2f}= K_{1i}+ K_{2i}+ U_{1i}+ U_{2i}( ^{1}/_{2}) (2 kg) v^{2}+ (^{1}/_{2}) (3 kg) v^{2}+ (2 kg)(9.8 m/s^{2})(2 m) + (3 kg)(9.8 m/s^{2})(2 m) == 0 + 0 + 0 + (3.0 kg) (9.8 m/s

^{2}) (4.0 m)( ^{1}/_{2}) (5 kg) v^{2}+ 98 J = 117.6 J(

^{1}/_{2}) (5 kg) v^{2}= 19.6 Jv

^{2}= (19.6 / 2.5)^{J}/_{kg}v

^{2}= 7.84 m^{2}/ s^{2}

v = 2.8 m / s

3. A toy consists of a piece of plastic attached to a spring as shown here. The spring is compressed 0.025 m and the toy is released. If the mass of the toy is 0.120 kg and it rises to a maximum height of 0.75 m, estimate the force constant of the spring. Initially, there is only spring potential energy stored in the spring,E _{i}= U_{si}= (^{1}/_{2}) k x^{2}= (^{1}/_{2}) k (0.025 m)^{2}Later, at the top of its "jump", when its velocity is momentarily zero, there is only gravitational potential energy,

E _{f}= U_{gf}= m g h_{f}= (0.120 kg) (9.8 m/s^{2}) (0.75 m) = 0.882 JSince we expect energy to be conserved, this means

E _{i}= E_{f}(

^{1}/_{2}) k (0.025 m)^{2}= 0.882 Jk = 2 822 N / m k = 2 800 N / m

4. The launcher from a pin ball machine has a spring constant of k = 150 N/m. A pin ball, with mass of m = 0.10 kg, is launched by pulling back the spring and compressing it a distance of x = 0.05 m. The ball is then launched up a 20° incline as shown above. How far along this incline does the ball travel before itstops. Neglect friction. Neglect the mass of the spring and launcher. Neglect the rotational effects of the pinball.When the launcher spring is compressed 0.050 m, the 0.10 kg pinball is

lowerby 0.017 m so its gravitational potential energy is lower, too. We will take this as the reference position from which we measure the gravitational potential energy. Then,U _{gi}= 0K

_{i}= 0U

_{si}= (^{1}/_{2}) k x^{2}U

_{si}= (^{1}/_{2}) (150^{N}/_{m}) (0.05 m)^{2}U

_{si}= 0.1875 JE = K + U

_{g}+ U_{s}E

_{i}= K_{i}+ U_{gi}+ U_{si}E

_{i}= 0 + 0 + 0.1875 JE

_{i}= 0.1875 JThe spring is released, so U

_{gf}= 0, and the pinball moves up the incline. Eventually it comes to rest, so K_{f}= 0. Where does this happen? It will be easier for us to calculate this in terms of the vertical distance y_{max}but it will be easier to describe or measure in terms of s, the distance along the incline. At this position,K _{f}= 0U

_{sf}= 0U

_{gf}= m g y_{max}E = K + U

_{g}+ U_{s}E

_{f}= K_{f}+ U_{gf}+ U_{s}fE

_{f}= 0 + m g y_{max }+ 0E

_{f}= E_{i}m g y

_{max}= 0.1875 J(0.10 kg) (9.8 m/s

^{2}) y_{max}= 0.1875 Jy

_{max}= 0.191 my

_{max}= (s) sin 20^{o}s = y

_{max}/ sin 20^{o}s = 0.191 m / 0.342

s = 0.56 m

Remember, we have measured all our distances from where the pinball is when the launcher spring is compressed. It has moved 0.05 m from its initial position. So, as it moves up along the incline a distance of 0.56 m, it has moved 0.51 m from its initial position, before the launcher spring is compressed;

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