Problem#1
Two forces F1 and F2 act on a 5.00-kg object. If F1 = 20.0 N and F2 = 15.0 N, find the accelerations in (a) and (b) of Figure 1.
Answer:
(a) the accelerations giben by
a = ∑F/m
with
∑F = F1 + F2 = (20.0i + 15.0j) N, then
a = (20.0i + 15.0j) N/(5.00 kg)
a =(4.0i + 3.0j) m/s2
or
a = √(4.02 + 3.02) = 5.0 m/s2
The direction is
tan θ = ay/ax = ¾
θ = 36.90
(b) F2x = 15.0 cos 60.00 = 7.50 N
F2y = 15.0 sin 60.00 = 13.0 N
F2 = F2xi + F2yj = (7.50i + 13.0j) N
Then
∑F = F1 + F2 = (27.5i + 13.0j) N = ma = 5.00 kg x a
a = (5.50i + 2.60j) m/s2
or
a = 6.08 m/s2 at 25.30
Problem#2
Besides its weight, a 2.80-kg object is subjected to one other constant force. The object starts from rest and in 1.20 s experiences a displacement of (4.20i + 3.30j) m, where the direction of j is the upward vertical direction. Determine the other force.
Answer;
We find acceleration:
rf = ri + vit + ½ at2
4.20 mi = 3.30 mj + ) + ½ a(1.20 s)2 = 0.720 s2a
a = (5.83i – 4.58j) m/s2
Now ∑F = ma becomes
Fg + F2 = ma
F2 = (2.80 kg)(5.83i – 4.58j) m/s2 + (2.80 kg)(9.80 m/s2)j
F2 = (16.3i + 14.6j) N
Problem#3
You stand on the seat of a chair and then hop off. (a) During the time you are in flight down to the floor, the Earth is lurching up toward you with an acceleration of what order of magnitude? In your solution explain your logic. Model the Earth as a perfectly solid object. (b) The Earth moves up through a distance of what order of magnitude?
Answer;
(a) You and the earth exert equal forces on each other:
myg = Meae
If your mass is 70.0 kg,
ae = (70.0 kg)(9.80 m/s2)/(5.98 x 1024 kg) = 10-22 m/s2
(b) You and the planet move for equal times intervals according to x= ½ at2. If the seat is
50.0 cm high,
[2xy/ay]1/2 = [2xe/ae]1/2
xe = (ae/ay)xy = (my/me)y
xe = (70.0 kg x 5.00 m)/(5.98 x 1024 kg) = 10-23 m
Problem#4
Three forces acting on an object are given by F1 = (–2.00i + 2.00j) N, F2 = (5.00i – 3.00j) N, and F3 = (–45.0i) N. The object experiences an acceleration of magnitude 3.75 m/s2. (a) What is the direction of the acceleration? (b) What is the mass of the object? (c) If the object is initially at rest, what is its speed after 10.0 s? (d) What are the velocity components of the object after 10.0 s?
Answer:
∑F = F1 + F2 + F3 = –2.00i + 2.00j + 5.00i – 3.00j – 45.0i
∑F = (–42.0i – 1.00j) N
Or
∑F = 42.0 N
the direction of the force given by
tan θ = (1/42)
θ = tan-1(1/42) = 1810 below the –x-axis
For the vectors to be equal, their magnitudes and their directions must be equal.
(a) so, the direction of acceleration is 1810 below the –x-axis
(b) we use
∑F = ma
Then
m = ∑F/a = 42.0 N/(3.75 m/s2) = 11.2 kg
(d) vf = vi + at = 0 + (3.75 m/s2 at 1810)(10.0 s)
So, vf =37.5 m/s at 1810 and
vf = 37.5 m/s cos 1800 i + 37.5 m/s sin 1810 j = (–37.5i – 0.893j) m/s
(c) |vf| = √[((–37.5)2 + (– 0.893)] = 37.5 m/s
Problem#5
A 15.0-lb block rests on the floor. (a) What force does the floor exert on the block? (b) If a rope is tied to the block and run vertically over a pulley, and the other end is attached to a free-hanging 10.0-lb weight, what is the force exerted by the floor on the 15.0-lb block? (c) If we replace the 10.0-lb weight in part (b) with a 20.0-lb weight, what is the force exerted by the floor on the 15.0-lb block?
Answer:
(a) force does the floor exert on the block is 15.0 lb up
(b) If a rope is tied to the block and run vertically over a pulley, and the other end is attached to a free-hanging 10.0-lb weight, the force exerted by the floor on the 15.0-lb block is
15.0 lb – 10.0 lb = 5.0 lb
(c) If we replace the 10.0-lb weight in part (b) with a 20.0-lb weight, the force exerted by the floor on the 15.0-lb block is zero.
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