PHY 1151
Spring 2000
Second Hour Exam
March 6, 2000

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Statistics:

High --> 100

Mean --> 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 showing all the forces on each of the blocks.

b) Find the acceleration of the blocks
Begin by applying Newton's Second Law, F = m a, to each of the blocks:

Look at the forces on the 2-kg hanging block --

F_net = w - T = ma

m g - T = m a

(2 kg)(10 m/s²) - T = (2 kg) a

20 N - T = (2 kg) a

That provides one equation but we have two unknowns (a and T). 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 = 0

F_N = w = mg = (5 kg)(10 m/s²)

F_N = 50 N

Therefore,

F_f = F_N

F_f = (0.2)(50 N) = 10 N

Now look at the forces on the 5-kg block that act in the horizontal direction,

F_{net, x} = T - F_f = m a

T - 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/s²

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/s²)

T = (20 - 2.9) kg m/s²

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 have F_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 is also the centripetal force F_c which points toward the center of the circle,

F_c = m v² / r

F_net = F_c

m g - F_N = m v² / r

F_N = m g - m v² / r

F_N, the normal force, is the force the road exerts on the car,

F_N = (1000 kg)(9.8 m/s²) - (1000 kg)(15 m/s)²/(40 m)

F_N = 9,800 N - 5,625 N

F_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 x²

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_f² = 0.25 J + 0

(¹/₂) m v_f² = 0.25 J

(¹/₂) (0.200 kg) v_f² = 0.25 J

v_f² = 2.5 m²/s²

v_f = 1.58 m/s

4. Consider a totally inelastic 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 is always conserved, P_f = P_i

P_f = (m₁ + m₂) v_f = m₁ v_1i + m₂ v_2i = P_i

(m₁ + m₂) v_f = (0.250 kg)(0.50 m/s) + (0.200 kg)( - 0.35 m/s)

(m₁ + m₂) v_f = [(0.250)(0.50) - (0.200)(0.35)] kg m/s

(m₁ + m₂) v_f = [0.125 - 0.070] kg m/s

(m₁ + m₂) v_f = 0.055 kg m/s

(0.450 kg) v_f = 0.055 kg m/s

v_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 = 0

F_net = F = T + T_r + w = 0

But, remember, these forces are all vectors. Any vector equation is elegant shorthand notation for two scalar equations.

	T + Tr + w = 0
Tx + Trx + wx = 0	and	Ty + Try + wy = 0
T - 0.500 Tr = 0		0 + 0.866 Tr - 20 N = 0
T = 0.500 Tr		0.866 Tr = 20 N
		Tr = 20 N / 0.866
T = 0.500 (23.1 N)		Tr = 23.1 N
T = 11.5 N