Exercise Solution
Problem 9
Part a.
Select up as positive.
Newton's law for the vertical direction gives
åFy
=
FL - Mg = Ma
ð
FL
= Mg + ma = M(g + .1g) = M(1.1 g) = 1.1Mg = 1.1M(9.8)
FL = 1.1 Mg
Part b.
The lifting force and displacement are in same direction so work done in
lifting a vertical distance h
WL
= FL cos00 = 1.10 Mgh
Problem 11
Work is equal to the area under the graph. It is roughly trapezodal, so
W = (1/2)(Fmax - Fmin)(db - da) = (1/2)250 N + 150 N)(35.0 m - 10.0 m)
W = 5,000 J
Problem 16
a.
KE = (1/2)mv2 so v = (2KE/m)1/2
so, if KE is doubled, V will be multiplied by a factor of (2)1/2
b.
KE = (1/2)mv2 so if v is doubled, then
KE will be
multiplied by a factor of 4
Problem 17
The work on the electron is equal to its change in KE
W = D KE = - 1.64 x 10-18 J
Problem 26
The elastic PE is given by
PEelastic = 1/2 kx2 x is the distance in compressing or stretching the string from its natural length.
x =( 2 PEelastic /k)1/2 = 0.34 m
x = 0.34 m
Problem 35
The forces acting on the sled are the force of gravity and the normal force. Draw a free body diagram as before. It can be seen that the normal force is perpendicular to the motion and therefore does no work.
The sled's mechanical energy is, therefore, conserved.
In the following, subscript 1 represents the sled at the bottom of the hill and subscript 2 is for the sled at the top of the hill.
Let the ground be the 0 position for PE (y = 0).
Knowns: y1 = 0, v2 = 0, y2 = 1.35 m
Using conservation of mechanical energy, we solve for v1, the speed at the bottom of the hill.
ME1 + PE1 = ME2 + PE2
(1/2)mv12 + 0 = 0 + mgy2
v1 = 5.14 m/s
Problem 38
Use conservation of energy. Subscript 1 represents the projectile at the launch point, and subscript 2 represents the projectile as it reaches the ground. The ground is the zero location for PE (y = 0). We have v1 = 185 m/s, y1 = 265 m, y2 = 0. Solve for v2.
v2
= (v12 + 2gy1)1/2
= 199 m/s
Problem 48
We can apply the conservation of energy to the child, using work done by gravity and work changed into thermal energy.
Subscript 1 is for the child at the top of the slide and 2 is for the child at the bottom. Let the ground be the 0 point for potential energy.
Known
v1 = 0 y1 = 3.5 m v2 = 2.2 m/s y2 = 0
Solve for work changed into thermal energy
E1 = E2 gives KE1 + PE1 = KE2 + PE2 +Wthermal
Wthermal = mgy1 - 1/2 mv22 gives Wthermal = 6.9 x 102 J
Problem 58
The work to lift the piano is the work done by the upward force. This force is equal in magnitude to the weight of the piano.
P = E/t = mgh/t
t = mgh/P (315 kg)(9.80 m/s2)(16.0 m)/1750 W = 28.2 s
t = 28.2 s