Elements of Physics, 50:750:131 Name ________________
Test 1 October 6, 2004
The working for all solutions must be shown. No credit will be given for an answer with no working. Take g to be 9.80 m/s2.
1. A cord is a volume of cut wood equal to a stack 8 ft long, 4 ft wide, and 4 ft high. A foot is 0.3048 m. How many cords are in 1.0 m3?
so the number of cords per cubic meter is
1/3.6246 = 0.27590
2. On a dry road, a car with good tires may be able to brake with a constant deceleration of 4.86 m/s2. (a) How long does such a car, initially traveling at
25.2 m/s, take to stop? (b) How far does it move in this time?
(a) Use v = v0 + at, with the final velocity of zero, the initial velocity 25.2 m/s and the acceleration -4.86 m/s2.
0 = 25.2m/s - (4.86 m/s2)t
t = 25.2/4.86 = 5.1852 s
(b) Use x-x0 = v0t + ½ at2
= (25.2m/s)(5.1852s) - 0.5 * (4.86 m/s2)(5.1852s)2
= 65.333 m
3. A stone is dropped into a river from a bridge 45.0 ft above the water. Another stone is thrown vertically down 1.2 s after the first is dropped. The two stones strike the water at the same time. (a) What is the initial speed of the second stone? (b) Sketch velocity versus time on a graph for each stone, taking zero time as the instant the first stone is released.
(I intended to type the height of the bridge as 45.0 METERS. If you took it to be this, I marked your answers as correct also)
(a) The time taken by the first stone is found from
y - y0 = v0yt - ½ gt2
0 - (45.0 * 0.3048) = 0 - ½ (9.8)t2
t2 = 2*45.0*0.3048 / 9.8 = 2.7992 s2
t = 1.6731 s
Since the second stone was dropped 1.2s later, its time in the air is 1.6731 - 1.2 = 0.4731 s.
To fall in this time, its initial speed is found using the same equation but with this time:
y - y0 = v0yt - ½ gt2
0 - (45.0 * 0.3048) = v0y(0.4731s) - ½ (9.8)(0.4731s)2
½ (9.8)(0.4731s)2 - (45.0 * 0.3048) = v0y(0.4731s)
v0y = -26.674 m/s
The minus sign shows that it was thrown downwards, its speed is 26.674 m/s.
The total x-component of the displacement is 70 + 0 + 25 sin(40°) = 86.070 km
The total y-component is 0 + 40 + 25 cos(40°) = 59.151 km
The magnitude of the total displacement is
The angle with the x-axis is given by tanq = y-component/x-component
q = 34.499°
5. A boat is traveling upstream in the positive direction of an x axis at 12 km/h with respect to the water of a river. The water is flowing at 9.0 km/h with respect to the ground. What are (a) the magnitude and (b) the direction of the boat's velocity with respect to the ground? A child on the boat walks from front to rear at 5.0 km/h with respect to the boat. What are the (c) magnitude and (d) direction of the child's velocity with respect to the ground?
(a) vBOAT-GROUND = vBOAT-WATER + vWATER-GROUND
= 12 km/h - 9 km/h (because the water is flowing in the negative direction
= 3 km/h
(b) the boat is moving in the positive direction with respect to the ground (because its v is positive)
(c) vCHILD-GROUND = vCHILD-BOAT + vBOAT-GROUND
= -5.0 km/h + 3 km/h
= -2.0 km/h
The magnitude of the child's velocity is 2.0 km/h, and (d) the direction is actually negative. Although the boat is moving upstream, the child's velocity relative to the boat is sufficiently negative that the child is moving downstream.
6. Three connected blocks are pulled to the right on a horizontal frictionless table by a force of magnitude T3 = 72.0 N. If m1 = 10.0 kg,
m2 = 12.0 kg, and m3 = 14 kg, calculate (a) the magnitude of the system's acceleration, (b) the tension T1, and (c) the tension T2.
(a) Choose all three blocks as the system. The mass of the system is then 10.0 + 12.0 + 14.) = 34 kg, and the only external horizontal force acting on it is the tension T3
Fnet = ma
72.0 N = (35 kg) a
a = 72 N / 36 kg = 2 m/s2
This is the acceleration for each of the blocks.
(b) Choose the mass 1 as the system. Now the only horizontal force acting is T1, so
T1 = m1a = (10 kg)(2 m/s2) = 20 N
(c) Choose 1 and 2 as the system
T2 = (10 kg + 12 kg)(2 m/s2) = 44 N
(Of course there are vertical forces, weights and normal forces. However, the blocks do not float up into the air or sink into the table, so their vertical accelerations are zero and all the vertical forces must cancel out.)
7. A 70 kg crate is dragged across a floor by pulling on a rope attached to the crate and inclined 30° above the horizontal. (a) If the coefficient of static friction is 0.50, what minimum force magnitude is required from the rope to start the crate moving? (b) If k = 0.40, what is the magnitude of the initial acceleration of the crate?
F F cos 30 F sin 30
W 0 -mg
N 0 N
F -µSN 0
------- -------------- ----------------
Fnet F cos 30 - µSN F sin 30 - mg + N
(a) If the crate is on the point of sliding, its acceleration is zero, and each of the totals is zero. Solve the y-equation for N as
N = mg - F sin 30
Then substitute this in the x-equation to get
F cos 30 - µS(mg - F sin 30) = 0
F = µSmg / (cos 30 + µS sin 30) = 307.34 N
(b) When the crate begins to slide, the expression for the x-component of the force is the same except that µS is replaced by µk.
F cos 30 - µkN = ma
ma = F cos 30 - µk(mg - F sin 30)
With the same value of F, this gives a = 0.760 m/s2
8. A ball is thrown horizontally from a height of 20 m and hits the ground with a speed that is three times its initial speed. What is the initial speed?
Notice that the initial velocity is in the x direction, and the initial y-component of the velocity is zero. Call the initial speed v0x. The ball falls through a vertical distance of 20 m. Its vertical velocity when it hits the ground is given by
The total speed is