Problem#1
A 70.0-kg diver steps off a 10.0-m tower and drops straight down into the water. If he comes to rest 5.00 m beneath the surface of the water, determine the average resistance force exerted by the water on the diver.Answer:
 |
| Fig.1 |
We shall take the zero level of gravitational potential energy to be at the lowest level reached by the
diver under the water, and consider the energy change from when the diver started to fall until he came to rest.
Then, given: yf = 0, yi = 10.0 m + 5.00 m = 15.0 m and m = 70.0 kg.
We use,
Wnc = ∆K + ∆U
f∆r cos1800 = 0 + mg(yf - yi)
–f(5.00 m) = (70.0 kg)(0 –15.0 m)
f = 2.06 x 103 N = 2.06 kN, pointing upward
Problem#2
The coefficient of friction between the 3.00-kg block and the surface in Figure 2 is 0.400. The system starts from rest. What is the speed of the 5.00-kg ball when it has fallen 1.50 m?
Answer:
Given: m1 = 3.00 kg, m2 = 5.00 kg, µ = 0.400, and d = 1.50 m.
Then, we use
Wnc = ∆K + ∆U
fd cos1800 = Kf – Ki + Uf – Ui
–µm1gd = ½(m1 + m2)v2 + 0 – m2gd
2(m2 – µm1)gd = (m1 + m2)v2
v = {2(m2 – µm1)gd/(m1 + m2)}1/2
v = {2(5.00 kg – 0.400(3.00 kg))(9.80 m/s2)(1.50 m)/(3.00 kg + 5.00 m)}1/2
v = 3.74 m/s
Problem#3
A boy in a wheelchair (total mass 47.0 kg) wins a race with a skateboarder. The boy has speed 1.40 m/s at the crest of a slope 2.60 m high and 12.4 m long. At the bottom of the slope his speed is 6.20 m/s. If air resistance and rolling resistance can be modeled as a constant friction force of 41.0 N, find the work he did in pushing forward on his wheels during the downhill ride.
Answer:
diver under the water, and consider the energy change from when the diver started to fall until he came to rest.
Then, given: yf = 0, yi = 10.0 m + 5.00 m = 15.0 m and m = 70.0 kg.
We use,
Wnc = ∆K + ∆U
f∆r cos1800 = 0 + mg(yf - yi)
–f(5.00 m) = (70.0 kg)(0 –15.0 m)
f = 2.06 x 103 N = 2.06 kN, pointing upward
Problem#2
The coefficient of friction between the 3.00-kg block and the surface in Figure 2 is 0.400. The system starts from rest. What is the speed of the 5.00-kg ball when it has fallen 1.50 m?
 |
| Fig.2 |
Given: m1 = 3.00 kg, m2 = 5.00 kg, µ = 0.400, and d = 1.50 m.
Then, we use
Wnc = ∆K + ∆U
fd cos1800 = Kf – Ki + Uf – Ui
–µm1gd = ½(m1 + m2)v2 + 0 – m2gd
2(m2 – µm1)gd = (m1 + m2)v2
v = {2(m2 – µm1)gd/(m1 + m2)}1/2
v = {2(5.00 kg – 0.400(3.00 kg))(9.80 m/s2)(1.50 m)/(3.00 kg + 5.00 m)}1/2
v = 3.74 m/s
Problem#3
A boy in a wheelchair (total mass 47.0 kg) wins a race with a skateboarder. The boy has speed 1.40 m/s at the crest of a slope 2.60 m high and 12.4 m long. At the bottom of the slope his speed is 6.20 m/s. If air resistance and rolling resistance can be modeled as a constant friction force of 41.0 N, find the work he did in pushing forward on his wheels during the downhill ride.
Answer:
 |
| Fig.3 |
we use
∆Emech = ∆K + ∆U
But, ∆Emech = Wapp – f∆x
where Wapp is the work the boy did pushing forward on the wheels.
Thus, Wapp – f∆x = ∆K + ∆U
Wapp = Kf – Ki + Uf – Ui + f∆x
Wapp = ½ m(vf2 – vi2) + mg(–h) + f∆x
Wapp = ½ (47.0 kg)[(6.20 m/s)2 – (1.40 m/s)2] – (47.0 kg)(9.80 m/s2)(2.60 m) + (41.0 N)(12.4 m)
Wapp = 168 J
Problem#4
A 5.00-kg block is set into motion up an inclined plane with an initial speed of 8.00 m/s (Fig. 4). The block comes to rest after traveling 3.00 m along the plane, which is inclined at an angle of 30.0° to the horizontal. For this motion determine (a) the change in the block’s kinetic energy, (b) the change in the potential energy of the block–Earth system, and (c) the friction force exerted on the block (assumed to be constant). (d) What is the coefficient of kinetic friction?
Answer:
(a) the change in the block’s kinetic energy, we use
∆K = ½ m(vf2 – vi2) = ½ (5.00 kg)[0 – (8.00 m/s)2) = –160 J
(b) the change in the potential energy of the block–Earth system,
∆U = mgd sin30.00 = (5.00 kg)(9.80 m/s2)(3.00 m)(0.500) = 73.5 J
(c) the friction force exerted on the block (assumed to be constant), we use
∆K + ∆U = Wnc = fd cos1800
–160 J + 73.5 J = f(3.00 m)(–1)
f = 86.5 J/3.00 m = 28.8 N
(d) the coefficient of kinetic friction is
f = µmg cos30.00
28.8 N = µ(50.0 N)cos30.00
µ = 0.679
Problem#5
An 80.0-kg skydiver jumps out of a balloon at an altitude of 1 000 m and opens the parachute at an altitude of 200 m. (a) Assuming that the total retarding force on the diver is constant at 50.0 N with the parachute closed and constant at 3 600 N with the parachute open, what is the speed of the diver when he lands on the ground? (b) Do you think the skydiver will be injured? Explain. (c) At what height should the parachute be opened so that the final speed of the skydiver when he hits the ground is 5.00 m/s? (d) How realistic is the assumption that the total retarding force is constant? Explain.
Answer:
(a) the speed of the diver when he lands on the ground with consider the whole motion, we use
–f1d1 – f2d2 = Kf – Ki + Uf – Ui
–(50.0 N)(800 m) – (3600 N)(200 m) = ½ (80.0 kg)vf2 – 0 + 0 – (80.0 kg)(9.80 m/s2)(1000 m)
–40000J – 720000 J = 40.0kgv2 – 784000J
24000 J = 40.0kgvf2
vf = 24.5 m/s
(b) Yes, this is too fast for safety.
(c) Now in the same energy equation as in part (a), d2 is unknown and d1 = 1000 m – d2:
–f1d1 – f2d2 = Kf – Ki + Uf – Ui
–(50.0 N)(1000 m – d2) – (3600 N)(d2) = ½ (80.0 kg)(5.00 m/s)2 – 0 + 0 – (80.0 kg)(9.80 m/s2)(1000 m)
–50000J – (3550 N)d2 = 1000 J – 784000J
d2 = 733000 J/3550 N = 206 m
But, ∆Emech = Wapp – f∆x
where Wapp is the work the boy did pushing forward on the wheels.
Thus, Wapp – f∆x = ∆K + ∆U
Wapp = Kf – Ki + Uf – Ui + f∆x
Wapp = ½ m(vf2 – vi2) + mg(–h) + f∆x
Wapp = ½ (47.0 kg)[(6.20 m/s)2 – (1.40 m/s)2] – (47.0 kg)(9.80 m/s2)(2.60 m) + (41.0 N)(12.4 m)
Wapp = 168 J
Problem#4
A 5.00-kg block is set into motion up an inclined plane with an initial speed of 8.00 m/s (Fig. 4). The block comes to rest after traveling 3.00 m along the plane, which is inclined at an angle of 30.0° to the horizontal. For this motion determine (a) the change in the block’s kinetic energy, (b) the change in the potential energy of the block–Earth system, and (c) the friction force exerted on the block (assumed to be constant). (d) What is the coefficient of kinetic friction?
 |
| Fig.4 |
(a) the change in the block’s kinetic energy, we use
∆K = ½ m(vf2 – vi2) = ½ (5.00 kg)[0 – (8.00 m/s)2) = –160 J
(b) the change in the potential energy of the block–Earth system,
∆U = mgd sin30.00 = (5.00 kg)(9.80 m/s2)(3.00 m)(0.500) = 73.5 J
(c) the friction force exerted on the block (assumed to be constant), we use
∆K + ∆U = Wnc = fd cos1800
–160 J + 73.5 J = f(3.00 m)(–1)
f = 86.5 J/3.00 m = 28.8 N
(d) the coefficient of kinetic friction is
f = µmg cos30.00
28.8 N = µ(50.0 N)cos30.00
µ = 0.679
Problem#5
An 80.0-kg skydiver jumps out of a balloon at an altitude of 1 000 m and opens the parachute at an altitude of 200 m. (a) Assuming that the total retarding force on the diver is constant at 50.0 N with the parachute closed and constant at 3 600 N with the parachute open, what is the speed of the diver when he lands on the ground? (b) Do you think the skydiver will be injured? Explain. (c) At what height should the parachute be opened so that the final speed of the skydiver when he hits the ground is 5.00 m/s? (d) How realistic is the assumption that the total retarding force is constant? Explain.
Answer:
(a) the speed of the diver when he lands on the ground with consider the whole motion, we use
Wnc = ∆K + ∆U
–f1d1 – f2d2 = Kf – Ki + Uf – Ui
–(50.0 N)(800 m) – (3600 N)(200 m) = ½ (80.0 kg)vf2 – 0 + 0 – (80.0 kg)(9.80 m/s2)(1000 m)
–40000J – 720000 J = 40.0kgv2 – 784000J
24000 J = 40.0kgvf2
vf = 24.5 m/s
(b) Yes, this is too fast for safety.
(c) Now in the same energy equation as in part (a), d2 is unknown and d1 = 1000 m – d2:
–f1d1 – f2d2 = Kf – Ki + Uf – Ui
–(50.0 N)(1000 m – d2) – (3600 N)(d2) = ½ (80.0 kg)(5.00 m/s)2 – 0 + 0 – (80.0 kg)(9.80 m/s2)(1000 m)
–50000J – (3550 N)d2 = 1000 J – 784000J
d2 = 733000 J/3550 N = 206 m
 |
| Fig.4 |
(d) Really the air drag will depend on the skydiver’s speed. It will be larger than her 784 N weight only after the chute is opened. It will be nearly equal to 784 N before she opens the chute and again before she touches down, whenever she moves near terminal speed.
Post a Comment for "Changes in Mechanical Energy for Nonconservative Forces Problems and Solutions 2"