Problem#1
Review problem. Two constant forces act on a 5.00-kg object moving in the xy plane, as shown in Figure 1. Force F1 is 25.0 N at 35.0°, while F2 is 42.0 N at 150°. At time t = 0, the object is at the origin and has velocity (4.00i + 2.50j) m/s. (a) Express the two forces in unitvector notation. Use unit-vector notation for your other answers. (b) Find the total force on the object. (c) Find the object’s acceleration. Now, considering the instant t = 3.00 s, (d) find the object’s velocity, (e) its location, (f) its kinetic energy from mvf2, and (g) its kinetic energy from ½ mvi2 + ∑F.∆r.
Answer:
(a) Express the two forces in unitvector notation. Use unit-vector notation for your other answers. We use
F = Fxi + Fyj
Then
F1 = (25.0 N)(cos35.00i + sin35.00j) = (20.5i + 14.3j) N, and
F2 = (42.0 N)(cos1500i + sin1500j) = (–36.4i + 21.0j) N
(b) the total force on the object is
F = F1 + F2 = (20.5i + 14.3j) N + (–36.4i + 21.0j) N
F = (–15.9i + 35.3j)N
(c) the object’s acceleration is
a = ∑F/m
a = (–15.9i + 35.3j)N/5.00kg = (–3.18i + 7.07j)m/s2
(d) the object’s velocity is
vf = vi + at
vf = (4.00i + 2.50j) m/s + (–3.18i + 7.07j)(m/s2) (3.00 s)
vf = (–5.54i + 23.7j)m/s
(e) its location, given by
rf = ri + vit + ½ at2
rf = 0 + (4.00i + 2.50j)(m/s)(3.00 s) + ½(–3.18i + 7.07j)(m/s2)(3.00 s)2
rf = (–2.30i + 39.3j)m
(f) its kinetic energy from mvf2, is
Kf = ½ (5.00 kg)[(–5.54 m/s)2 + (23.7 m/s)2] = 1.48 kJ
(g) its kinetic energy from ½ mvi2 + ∑F.∆r, is
Kf = ½ (5.00 kg)[(4.00 m/s)2 + (2.50 m/s)2] + (–15.9i + 35.3j)N(–2.30i + 39.3j)m
Kf = 55.6 J + (–15.9N)(–2.30 m) + (35.3 N)(39.3 m)
Kf = 55.6J + 1426 J = 1.48 kJ
Problem#2
A 200-g block is pressed against a spring of force constant 1.40 kN/m until the block compresses the spring 10.0 cm. The spring rests at the bottom of a ramp inclined at 60.0° to the horizontal. Using energy considerations, determine how far up the incline the block moves before it stops (a) if there is no friction between the block and the ramp and (b) if the coefficient of kinetic friction is 0.400.
Answer:
(a) We use, ∑W = ∆K + ∆Eint, but ∆K = 0 and ∆Eint = 0
∑W = 0
Ws + Wg = 0
½ kxi2 – 0 + mg∆xcos(900 + 600) = 0
½ (1.40 x 103N/m)(0.100 m)2 + (0.200 kg)(9.80 m/s2) (–½√3)∆x = 0
∆x = 4.12 m
(a) We use, ∑W = ∆K + ∆Eint, but ∆K = 0 and ∆Eint = 0
∑W = 0
Ws + Wg – ∆Eint = 0
½ kxi2 – 0 + mg∆xcos(900 + 600) – µmg∆xcos600 = 0
½ (1.40 x 103N/m)(0.100 m)2 + (0.200 kg)(9.80 m/s2)(–½√3)∆x + (0.400)(0.200 kg)(9.80 m/s2)(½√3)∆x = 0
∆x = 3.35 m
Problem#3
When different weights are hung on a spring, the spring stretches to different lengths as shown in the following table. (a) Make a graph of the applied force versus the extension of the spring. By least-squares fitting, determine the straight line that best fits the data. (You may not want to use all the data points.) (b) From the slope of the bestfit line, find the spring constant k. (c) If the spring is extended to 105 mm, what force does it exert on the suspended weight?
Answer:
(a)
(b) A straight line fits the first eight points, together with the origin. By least-square fitting, its
slope is
0.125 N/mm ± 2% = 125 N/m ± 2%
In F = kx, the spring constant is k = F/x, the same as the slope of the F-versus-x graph.
(c) force does it exert on the suspended weight if x = 105 mm is
F = kx = (125N/m)(0.105 m) = 13.2 N
Problem#5
The ball launcher in a pinball machine has a spring that has a force constant of 1.20 N/cm (Fig. 3). The surface on which the ball moves is inclined 10.0° with respect to the horizontal. If the spring is initially compressed 5.00 cm, find the launching speed of a 100-g ball when the plunger is released. Friction and the mass of the plunger are negligible.
Answer:
We use
∑W = ∆K
Ws + Wg = Kf – Ki
½ kxi2 – ½ kxf2 + mg∆x cos1000 = ½ mvf2 – ½ mvi2
½ kxi2 – 0 + mg∆x cos1000 = ½ mvf2 – 0
½ (1.20 N/cm)(5.00 cm)(0.0500 m) + (0.100kg)(9.80 m/s2)(0.0500 m)sin10.00 = ½ (0.100 kg)v2
0.150J – 8.51 x 10-3 J = (0.0500 kg)v2
v = 1.68 m/s
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