Download Honors Physics Name HW – Forces, F = ma, and Equilibrium Date

Honors Physics
_______________________
HW – Forces, F = ma, and Equilibrium
________________________
Name
Date
1. After a skydiver jumps from a plane, the only force initially acting on the diver is Earth’s
gravitational attraction. After about ten seconds of falling, air resistance on the diver
will have increased so that its magnitude on the diver is now equal in magnitude to
Earth’s gravitational force on the diver. At this time, a diver in belly-down position will
be falling at a constant speed of about 190 km/h.
a. Draw a free-body diagram of the skydiver when the diver initially leaves the plane.
b. Draw a free-body diagram of the skydiver at the tenth second of falling.
2. A chef places an open sack of flour on a kitchen scale. The scale reading of 40 N indicates
that the scale is exerting an upward force of 40 N on the sack. The magnitude of this
force equals the magnitude of the fore of Earth’s gravitational attraction on the sack. The
chef then exerts an upward force of 10 N on the bag and the scale reading falls to 30 N.
Draw a free body diagram of the situation with the chef exerting an upward force on the
sack of flour.
3. To the right is the picture of a child’s toy. The child pushes the toy with a force of 5 N
and the floor exerts a force of 13 N directly upward on the toy. The Earth’s gravitational
force on the toy is 10 N downward while interactions between the
wheels and the floor produce a backward force of 2 N on the toy as
is moves. Draw a free- body diagram of the toy as it is pushed.
1
4. Page 114 # 47
Net force only relates to the acceleration. If the net force is to the right then the acceleration is
to the right. If the motion is to the right then the object is speeding up (velocity and
acceleration are in the same direction). However, if the motion is to the left then the object is
slowing down (velocity and acceleration are in opposite directions). Thus, we are unable to
determine if the object is moving left or right without more information.
5. Page 114 # 51
a. Assuming that the winch is moving at a constant velocity (no acceleration), then the
winch can pull up a weight of 2000 N. To determine the mass, use the formula W =
mg.
𝑊 = 𝑚𝑔
𝑊
−2000 𝑁
𝑚=
=
= 204 𝑘𝑔 𝑤𝑖𝑡ℎ 𝑠𝑖𝑔 𝑓𝑖𝑔𝑠 𝑖𝑡 𝑖𝑠 200 𝑘𝑔
9.8𝑚
𝑔
− 2
𝑠
b. Fwinch = 1200 N
Fg = mg = (120kg) (-9.8 m/s2) = -1176 N
m = 120 kg
𝐹𝑛𝑒𝑡 = 𝐹𝑤𝑖𝑛𝑐ℎ + 𝐹𝑔 = 1200 𝑁 + −1176 𝑁 = 24 𝑁
𝐹𝑛𝑒𝑡 = 𝑚𝑎
𝐹𝑛𝑒𝑡
24 𝑁
𝑚
𝑎=
=
= 0.20 2
𝑚
120 𝑘𝑔
𝑠
c. A = 0.20 m/s2
Δy = 12m
vi = 0
t=?
1
Δ𝑦 = 𝑣𝑦𝑖 Δ𝑡 + 𝑎Δ𝑡 2
2
1
Δ𝑦 = 𝑎Δ𝑡 2 𝑏𝑒𝑐𝑎𝑢𝑠𝑒 𝑣𝑦𝑖 = 0
2
2Δ𝑦
2 (12𝑚)
Δ𝑡 = √
= √
𝑚 = 11 𝑠𝑒𝑐𝑜𝑛𝑑𝑠
𝑎
0.20 2
𝑠
6. Page 115 # 52
a. Constant speed so no acceleration. Normal force and force of gravity are balanced.
Thus, W = mg = (53 kg)(-9.8 m/s2) = -520 N. The normal force will be equal but
opposite sign. FN = 520 N.
b. Velocity is up. Acceleration is down at -2.0 m/s2.
𝐹𝑛𝑒𝑡 = 𝑚𝑎
𝑚
𝐹𝑛𝑒𝑡 = (53𝑘𝑔) (−2.0 2 ) = −106 𝑁
𝑠
𝐹𝑛𝑒𝑡 = 𝐹𝑔 + 𝐹𝑁
𝐹𝑁 = 𝐹𝑛𝑒𝑡 − 𝐹𝑔
𝐹𝑁 = −106𝑁 − (−520𝑁) = 414 𝑁
c. Velocity is down. Acceleration is down at -2.0 m/s2.
𝐹𝑛𝑒𝑡 = 𝑚𝑎
𝑚
𝐹𝑛𝑒𝑡 = (53𝑘𝑔) (−2.0 2 ) = −106 𝑁
𝑠
𝐹𝑛𝑒𝑡 = 𝐹𝑔 + 𝐹𝑁
𝐹𝑁 = 𝐹𝑛𝑒𝑡 − 𝐹𝑔
2
𝐹𝑁 = −106𝑁 − (−520𝑁) = 414 𝑁
d. Velocity is constant. Acceleration is zero. So, Fnet = 0 and Fg = - FN. FN = 520 N.
e. Velocity is down. Acceleration is up at 2.5 m/s2.
𝐹𝑛𝑒𝑡 = 𝑚𝑎
𝑚
𝐹𝑛𝑒𝑡 = (53𝑘𝑔) (2.5 2 ) = 132.5 𝑁
𝑠
𝐹𝑛𝑒𝑡 = 𝐹𝑔 + 𝐹𝑁
𝐹𝑁 = 𝐹𝑛𝑒𝑡 − 𝐹𝑔
𝐹𝑁 = 132.5 𝑁 − (−520𝑁) = 652 𝑁
7. Page 115 # 53
a. gmercury = (0.38) gEarth = (0.38) (-9.8 m/s2) = -3.724 m/s2
𝑊 = 𝑚 𝑔𝑚𝑒𝑟𝑐𝑢𝑟𝑦
𝑚
𝑊 = (6.0 𝑘𝑔) (−3.724 2 ) = −22 𝑁
𝑠
b. gpluto = (0.08) gmercury = (0.08) (-3.724 m/s2) = -0.29792 m/s2
𝑊 = 𝑚 𝑔𝑝𝑙𝑢𝑡𝑜
𝑚
𝑊 = (7.0 𝑘𝑔) (−0.29792 2 ) = −2.1 𝑁
𝑠
8. Page 115 # 61 – Assuming that it is falling with a constant velocity then Fraindrop = -Fg. Fg =
mg. 2.45 mg = 2.45 x 10-6 kg. Fg = (2.45 x 10-6 kg)(-9.8 m/s2) = -2.40 x 10-5 N. Raindrop
will have a force of 24.0 N.
9. Page 115 # 62
Lion
Sprinter
m = 170 kg
m = 75 kg
a = 10.0 m/s2
a = 10.0 m/s2
FN = -Fg = -mg = -(170 kg)(-9.8 m/s2) = 1666 N
FN = -Fg = -mg = -(75 kg)(-9.8 m/s2) = 735 N
Fnetx = max = (170 kg)( 10.0 m/s2) = 1700 N
Fnetx = max = (75 kg)( 10.0 m/s2) = 750 N
Difference in normal force = 931 N
Difference in forward force = 950 N
10. Page 115 # 64
Top box has a Fg = (4.6 kg)(-9.8 m/s2) = -45.08 N
Normal force is equal and opposite = 45 N
Middle bos box has a Fg = (4.6 + 1.2 kg)(-9.8 m/s2) = -56.84 N
Normal force is equal and opposite = 57 N
4.6 kg
1.2 kg
Bottom box has a Fg = (4.6 + 1.2 + 3.7 kg)(-9.8 m/s2) = -93.1 N
Normal force is equal and opposite = 93 N
3.7 kg
11. MOP NL #2 – codes are due on Tuesday, December 16 by 9 AM
12. MOP NL #3 – codes are due on Tuesday, December 16 by 9 AM
3