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Problems & Exercises


265 N


13.3 m/s 2 13.3 m/s 2


(a) 12 m/s 2 12 m/s 2 .

(b) The acceleration is not one-fourth of what it was with all rockets burning because the frictional force is still as large as it was with all rockets burning.


(a) The system is the child in the wagon plus the wagon.


An object represented as a dot labeled m is shown at the center. One force represented by an arrow labeled as vector F sub 2 acts toward the right. Another force represented by an arrow labeled as vector F sub 1 having a slightly shorter length in comparison with F sub 2 acts on the object pointing left. A friction force represented by an arrow labeled as vector f having a small length acts on the object toward the left. Weight, represented by an arrow labeled as vector W, acts on the object downward, and normal force, represented by an arrow labeled as vector N, acts upward, having the same length as W.

(c) a=0.130 m/s2a=0.130 m/s2 in the direction of the second child’s push.

(d) a=0.00 m/s2a=0.00 m/s2


(a) 3.68 × 10 3 N 3.68× 10 3 N . This force is 5.00 times greater than his weight.

(b) 3750 N; 11.3º above horizontal3750 N; 11.3º above horizontal


1.5 × 10 3 N , 150 kg , 150 kg 1.5× 10 3 N,150 kg,150 kg


Force on shell: 2.64×107N2.64×107N

Force exerted on ship = 2.64×107N2.64×107N, by Newton’s third law


(a) 0.11 m/s2 0.11 m/s2
(b) 1.2× 10 4 N1.2× 10 4 N


(a) 7.84×10-4 N7.84×10-4 N

(b) 1.89×10–3 N1.89×10–3 N. This is 2.41 times the tension in the vertical strand.


Newton’s second law applied in vertical direction gives

F y = F 2 T sin θ = 0 F y = F 2 T sin θ = 0
F = 2T sin θ F = 2T sin θ
T = F 2 sin θ . T = F 2 sin θ .
An object of mass m is shown. Three forces acting on it are tension T, shown by an arrow acting vertically upward, and friction f and gravity m g, shown by two arrows acting vertically downward.

Using the free-body diagram:


so that

a=Tfmgm=1.250×107 N4.50×106 N(5.00×105 kg)(9.80 m/s2)5.00×105 kg=6.20 m/s2a=Tfmgm=1.250×107 N4.50×106 N(5.00×105 kg)(9.80 m/s2)5.00×105 kg=6.20 m/s2.

  1. Use Newton’s laws of motion. Two forces are acting on an object of mass m: F, shown by an arrow pointing upward, and its weight w, shown by an arrow pointing downward. Acceleration a is represented by a vector arrow pointing upward. The figure depicts the forces acting on a high jumper.
  2. Given : a=4.00g=(4.00)(9.80 m/s2)=39.2 m/s2 ; a=4.00g=(4.00)(9.80 m/s2)=39.2 m/s2 ; m=70.0 kgm=70.0 kg,

    Find: FF.

  3. F=+Fw=ma ,F=+Fw=ma , so that F=ma+w=ma+mg=m(a+g)F=ma+w=ma+mg=m(a+g).

    F=(70.0 kg)[(39.2 m/s2)+(9.80 m/s2)]F=(70.0 kg)[(39.2 m/s2)+(9.80 m/s2)]=3.43×103 N=3.43×103 N. The force exerted by the high-jumper is actually down on the ground, but FF is up from the ground and makes him jump.

  4. This result is reasonable, since it is quite possible for a person to exert a force of the magnitude of 103 N103 N.

(a) 4.41×105 N4.41×105 N

(b) 1.50×105 N1.50×105 N


(a) 910 N910 N

(b) 1.11×103 N1.11×103 N


a = 0.139 m/sa=0.139 m/s, θ = 12.4º θ=12.4º north of east

  1. Use Newton’s laws since we are looking for forces.
  2. Draw a free-body diagram: A horizontal dotted line with two vectors extending downward from the mid-point of the dotted line, both at angles of fifteen degrees. A third vector points straight downward from the intersection of the first two angles, bisecting them; it is perpendicular to the dotted line.
  3. The tension is given as T=25.0 N.T=25.0 N. Find Fapp.Fapp. Using Newton’s laws gives: Σ Fy=0,Σ Fy=0, so that applied force is due to the y-components of the two tensions: Fapp=2 T sin θ=2(25.0 N)sin(15º)=12.9 NFapp=2 T sin θ=2(25.0 N)sin(15º)=12.9 N

    The x-components of the tension cancel. Fx=0Fx=0.

  4. This seems reasonable, since the applied tensions should be greater than the force applied to the tooth.

10.2 m/s 2 , 4.67º from vertical 10.2 m/s 2 , 4.67º from vertical

An object of mass m is shown being pulled by two ropes. Tension T sub two acts toward the right at an angle of ten degrees above the horizontal. Another rope makes an angle fifteen degrees to the left of the vertical direction, and tension in the rope is T sub one, shown by a vector arrow. Weight w is acting vertically downward.

T 1 = 736 N T 1 = 736 N

T 2 = 194 N T 2 = 194 N


(a) 7.43 m/s7.43 m/s

(b) 2.97 m


(a) 4.20 m/s4.20 m/s

(b) 29.4 m/s229.4 m/s2

(c) 4.31×103 N4.31×103 N


(a) 47.1 m/s

(b) 2.47×103 m/s22.47×103 m/s2

(c) 6.18×103 N6.18×103 N . The average force is 252 times the shell’s weight.


(a) 1×10131×1013

(b) 1×10111×1011


10 2 10 2


(a) Box A travels faster at the finishing distance since a greater force with equal mass results in a greater acceleration. Also, a greater acceleration over the same distance results in a greater final speed.

(b) i. Yes, it is consistent because a greater force results in a greater final speed. ii. It does not make sense because V= 2(f/m)x =K (F) V= 2(f/m)x =K (F) .


The figure is a plot of x as a function of F. The curve starts at a high value of x just to the right of the x axis and decreases in both value and slope, tending asymptotically to the F axis. Under the graph it is noted that x is proportional to 1 over F.
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