**PHY
1151**

Spring 2000

**Second Hour Exam**

March 6, 2000

## Statistics:

High --> 100Mean --> 61 (up from 51 on the First exam!)

Low --> 6

**1.** A 5-kg block sits on a horizontal plane.
The 5-kg block is attached to a string which runs over a pulley and
is then attached to a 2-kg hanging block. The coefficient of sliding
friction between the 5-kg block and the horizontal plane is
µ = 0.2.

a) Draw good, clear diagrams showingallthe forces on each of the blocks.b) Find the acceleration of the blocks

Begin by applying Newton's Second Law,F= ma, to each of the blocks:Look at the forces on the 2-kg hanging block --

F _{net}= w - T = mam g - T = m a

(2 kg)(10 m/s

^{2}) - T = (2 kg) a20 N - T = (2 kg) a

That provides

oneequationbut we havetwounknowns(aandT). So we must look at the forces on the 5-kg block on the plane. It has forces in both the vertical and horizontal directions. First, look at the forces in the vertical direction. This will allow us to solve for the normal force F_{N}which we will need to solve for the fricition force F_{f}F _{net, y}= F_{N}- w = m a_{y}F

_{N}- w = 0F

_{N}= w = mg = (5 kg)(10 m/s^{2})F

_{N}= 50 NTherefore,

F _{f}= F_{N}F

_{f}= (0.2)(50 N) = 10 NNow look at the forces on the 5-kg block that act in the

horizontaldirection,F _{net, x}= T - F_{f}= m aT - 10 N = (5 kg) a

Again, we now have two unknowns in one equation. We can go back to our earlier equation from the 2-kg hanging mass and substitute our expression for the tension T,

20 N - T = (2 kg) a T = 20 N - (2 kg) a

T - 10 N = (5 kg) a

[20 N - (2 kg) a] - 10 N = (5 kg) a

20 N - 10 N = (5 kg) a + (2 kg) a

10 N = (7 kg) a

a =

^{10 N }/_{ 7 kg}

a = 1.43 m/s^{2}c) Find the tension in the string.

We have already solved for the tension T in terms of the acceleration,T = 20 N - (2 kg) a T = 20 N - (2 kg) (1.43 m/s

^{2})T = (20 - 2.9) kg m/s

^{2}

T = 17.1 N

**2.** A car of mass 1000 kg goes over the
crest of a hill whose radius of curvature is 40 m measured in a
vertical plane.

What is the force of the road surface on the car if the car’s
speed is 15 m/s?

From the diagram we haveF _{net}= m g - F_{N}where we have taken "down" as "positive" since that is the direction toward the center of the circle. The net force F

_{net}isalsothe centripetal force F_{c}which points toward the center of the circle,F _{c}= m v^{2}/ rF

_{net}= F_{c}m g - F

_{N}= m v^{2}/ rF

_{N}= m g - m v^{2}/ rF

_{N}, thenormalforce, is the force the road exerts on the car,F _{N}= (1000 kg)(9.8 m/s^{2}) - (1000 kg)(15 m/s)^{2}/(40 m)F

_{N}= 9,800 N - 5,625 NF

_{N}= 9,800 N - 5,625 N

F_{N}= 4,175 N

**3**. At the end of an air track, a spring
with spring constant of 50 N/m is compressed 0.10 m.

a) How much potential energy is stored in the spring?PE = ( ^{1}/_{2}) k x^{2}PE = (

^{1}/_{2}) (50^{N}/_{m}) (0.10 m)^{2}

PE = 0.25 J

A 0.200-kg air track glider is placed next to the compressed spring. The spring is released. The glider looses contact with the spring when the spring returns to its equilibrium position.

b) What is the speed of the glider as it leaves the spring?E _{f}= E_{i}PE

_{f}+ KE_{f}= PE_{i}+ KE_{i}0 + (

^{1}/_{2}) m v_{f}^{2}= 0.25 J + 0(

^{1}/_{2}) m v_{f}^{2}= 0.25 J(

^{1}/_{2}) (0.200 kg) v_{f}^{2}= 0.25 Jv

_{f}^{2}= 2.5 m^{2}/s^{2}

v_{f}= 1.58 m/s

**4.** Consider a totally __in__elastic
collision between two gliders on an airtrack.

Glider #1 has a mass of 250 g (or 0.250 kg) and moves to the right
with an initial speed of 50 cm/s (or 0.50 m/s). Glider #2 has a mass
of 200 g (or 0.200 kg) and moves to the __left__ with an initial
speed of 35 cm/s (or 0.35 m/s).

The two collide and stick together and move off with a common
velocity, v_{f}.

Find that common, final velocity, v_{f}.

Momentum isalwaysconserved,P _{f}= P_{i}P

_{f}= (m_{1}+ m_{2}) v_{f}= m_{1}v_{1i}+ m_{2}v_{2i}= P_{i}(m

_{1}+ m_{2}) v_{f}= (0.250 kg)(0.50 m/s) + (0.200 kg)( - 0.35 m/s)(m

_{1}+ m_{2}) v_{f}= [(0.250)(0.50) - (0.200)(0.35)] kg m/s(m

_{1}+ m_{2}) v_{f}= [0.125 - 0.070] kg m/s(m

_{1}+ m_{2}) v_{f}= 0.055 kg m/s(0.450 kg) v

_{f}= 0.055 kg m/sv

_{f}= (0.055/0.450) m/s

v_{f}= 0.122 m/s

v_{f}=12.2 cm/s

_{
}

**5.** A 2-kg ball is held in position by a
horizontal string and a string that makes an angle of 30° with
the vertical, as shown in the figure. Find the tension T in the
horizontal string,.

The first condition of equilibrium is that the vector sum of all the forces on an object is zero. Look at the forces on the knot and set them equal to zero.F

_{net}=F= 0F

_{net}=F=T+T_{r}+w= 0But,

remember, these forces are allvectors. Any vector equation is elegant shorthand notation fortwoscalar equations.

T+T_{r}+w= 0T _{x}+ T_{rx}+ w_{x}= 0and T _{y}+ T_{ry}+ w_{y}= 0T - 0.500 T _{r}= 00 + 0.866 T _{r}- 20 N = 0T = 0.500 T _{r}0.866 T _{r}= 20 NT _{r}= 20 N / 0.866T = 0.500 (23.1 N) T _{r}= 23.1 NT=11.5 N

(c) 2000, Doug Davis; all rights reserved