1. REASONING The work done by the catapult catapult is one contribution to the work done by the net external force that changes the kinetic energy of the plane. The other contribution is the work done by the thrust force of the plane s engines thrust. According to the workenergy theorem (Equation 6.3), the work done by the net external force catapult + thrust is equal to the change in the kinetic energy. The change in the kinetic energy is the given kinetic energy of 4.5 10 7 J at lift-off minus the initial kinetic energy, which is zero since the plane starts at rest. The work done by the thrust force can be determined from Equation 6.1 [ = (F cos ) s], since the magnitude F of the thrust is.3 10 5 N and the magnitude s of the displacement is 87 m. e note that the angle between the thrust and the displacement is 0º, because they have the same direction. In summary, we will calculate catapult from catapult + thrust = KE f KE 0. SOLUTION According to the work-energy theorem, we have catapult + thrust = KE f KE 0 Using Equation 6.1 and noting that KE 0 = 0 J, we can write the work energy theorem as follows: + catapult ( Fcosθ ) s = KE f Solving for catapult gives ( θ ) = KEcos F s catapult f ork done by thrust ork done by thrust = = 7 5 7 4.5 10 J.3 10 N cos 0 87 m.5 10 J
15. REASONING AND SOLUTION a. The work-energy theorem gives = (1/)mv f (1/)mv o = (1/)(45 10 3 kg)(41 m/s) = 38J b. From the definition of work so = Fs cos 0 F = /s = (38 J)/(1.0 10 m) = 3.8 10 3 N
17. SSM REASONING AND SOLUTION The work required to bring each car up to speed is, from the work-energy theorem, = KE f KE 0 = 1 mv f 1 mv 0. Therefore, 1 ( ) 1 3 5 B = f 0 = m v v (1.0 10 kg) (40.0 m/s) 0 m/s = 9.60 10 J 1 ( ) 1 3 6 B = f 0 = m v v (.00 10 kg) (40.0 m/s) 0 m/s = 1.60 10 J The additional work required to bring car B up to speed is, therefore, B A = (1.6 10 6 J) (9.60 10 5 J) = 6.4 10 5 J
0. REASONING To find the coefficient of kinetic friction µ k, we need to find the force of kinetic friction f k and the normal force F N (see Equation 4.8, f k = µ k F N ). The normal force and the weight mg of the sled balance, since they are the only two forces acting vertically and the sled does not accelerate in the vertical direction. The force of kinetic friction can be obtained from the work f done by the frictional force, according to Equation 6.1 [ f = (f k cos ) s], where s is the magnitude of the displacement. To find the work, we will employ the work-energy theorem, as given in Equation 6.3 ( = KE f KE 0 ). In this equation is the work done by the net force, but the normal force and the weight balance, so the net force is that due to the pulling force P and the frictional force. As a result = pull + f. SOLUTION According to the work-energy theorem, we have = pull + f = KE f KE 0 Using Equation 6.1 [ = (F cos ) s] to express each work contribution, writing the kinetic 1 energy as mv, and noting that the initial kinetic energy is zero (the sled starts from rest), we obtain 1 ( P cos 0 ) s+ ( f cos180 k ) s = mv pull The angle between the force and the displacement is 0º for the pulling force (it points in the same direction as the displacement) and 180º for the frictional force (it points opposite to the displacement). Equation 4.8 indicates that the magnitude of the frictional force is f k = µ k F N, and we know that the magnitude of the normal force is F N = mg. ith these substitutions the work-energy theorem becomes 1 ( P cos 0 ) s+ ( µ mg cos180 ) s = mv pull k Solving for the coefficient of kinetic friction gives f f ( mg ) s 1 1 mv P cos 0 s 16 kg.0 m/s 4 N 8.0 m µ = = = k cos180 16 kg9.80 m/s 8.0 m 0.13
. REASONING Since the person has an upward acceleration, there must be a net force acting in the upward direction. The net force ΣF y is related to the acceleration a y by Newton s second law, Σ Fy = may, where m is the mass of the person. This relation will allow us to determine the tension in the cable. The work done by the tension and the person s weight can be found directly from the definition of work, Equation 6.1. SOLUTION a. The free-body diagram at the right shows the two forces that act on the person. Applying Newton s second law, we have T s +y T mg = ma Σ F y y Solving for the magnitude of the tension in the cable yields mg T = m(a y + g) = (79 kg)(0.70 m/s + 9.80 m/s ) = 8.3 10 N b. The work done by the tension in the cable is ( θ ) = T s= (6.1) 3 T cos (8.3 10 N) (cos 0 ) (11 m) = 9.1 10 J c. The work done by the person s weight is ( θ ) = mg s = (6.1) 3 cos (79 kg) 9.8 m/s (cos 180 ) (11 m) = 8.5 10 J d. The work-energy theorem relates the work done by the two forces to the change in the kinetic energy of the person. The work done by the two forces is = T + : 1 1 T + = mv f mv 0 (6.3) Solving this equation for the final speed of the person gives v = v + + m f 0 T 3 3 = ( 0 m/s) + ( 9.1 10 J 8.5 10 J) = 4 m / s 79 kg
3. REASONING It is useful to divide this problem into two parts. The first part involves the skier moving on the snow. e can use the work-energy theorem to find her speed when she comes to the edge of the cliff. In the second part she leaves the snow and falls freely toward the ground. e can again employ the work-energy theorem to find her speed just before she lands. SOLUTION The drawing at the right shows the three forces that act on the skier as she glides on the snow. The forces are: her weight mg, the normal force F N, and the kinetic frictional force f k. Her displacement is labeled as s. The work-energy theorem, Equation 6.3, is f k 65.0 F N s 5.0 +y = mv mv 1 1 f 0 where is the work done by the net external force that acts on the skier. The work done by each force is given by Equation 6.1, = F cos θ s, so the work-energy theorem becomes mg 1 1 ( mg cos 65.0 ) s + ( f cos 180 ) s + ( F cos 90 ) s = mv mv k N f 0 Since cos 90 = 0, the third term on the left side can be eliminated. The magnitude f k of the kinetic frictional force is given by Equation 4.8 as fk = µ kfn. The magnitude F N of the normal force can be determined by noting that the skier does not leave the surface of the slope, so a y = 0 m/s. Thus, we have that ΣF y = 0, so F mg cos 5.0 = 0 or F = mg cos 5.0 N ΣF y The magnitude of the kinetic frictional force becomes fk = µ kfn = µ kmg cos 5.0 Substituting this result into the work-energy theorem, we find that N. 1 1 ( mg cos 65.0 ) s + ( µ mg cos 5.0 )( cos 180 ) s = mv mv k f 0 Algebraically eliminating the mass m of the skier from every term, setting cos 180 = 1 and v 0 = 0 m/s, and solving for the final speed v f, gives f v = gs cos 65.0 µ cos 5.0 k = 9.80 m/s 10.4 m cos65.0 0.00 cos 5.0 = 7.01 m/s
The drawing at the right shows her displacement s during free fall. Note that the displacement is a vector that starts where she leaves the slope and ends where she touches the ground. The only force acting on her during the free fall is her weight mg. The work-energy theorem, Equation 6.3, is 3.50 m θ s mg = mv mv 1 1 f 0 The work is that done by her weight, so the work-energy theorem becomes 1 1 ( mg cosθ ) s = mv mv f 0 In this expression θ is the angle between her weight (which points vertically downward) and her displacement. Note from the drawing that s cos θ = 3.50 m. Algebraically eliminating the mass m of the skier from every term in the equation above and solving for the final speed v f gives f 0 v = v + cos g s θ = 7.01 m/s + 9.80 m/s 3.50 m = 10.9 m/s
6. REASONING The work done by the weight of the basketball is given by Equation 6.1 as = ( F cos θ ) s, where F = mg is the magnitude of the weight, θ is the angle between the weight and the displacement, and s is the magnitude of the displacement. The drawing shows that the weight and displacement are parallel, so that θ = 0. The potential energy of the basketball is given by Equation 6.5 as PE = mgh, whereh is the height of the ball above the ground. SOLUTION a. The work done by the weight of the basketball is mg s ( cos θ ) = F s = mg (cos 0 )(h 0 h f ) = (0.60 kg)(9.80 m/s )(6.1 m 1.5 m) = 7 J b. The potential energy of the ball, relative to the ground, when it is released is PE 0 = mgh 0 = (0.60 kg)(9.80 m/s )(6.1 m) = 36 J (6.5) c. The potential energy of the ball, relative to the ground, when it is caught is PE f = mgh f = (0.60 kg)(9.80 m/s )(1.5 m) = 8.8 J (6.5) d. The change in the ball s gravitational potential energy is PE = PE f PE 0 = 8.8 J 36 J = 7 J e see that the change in the gravitational potential energy is equal to 7 J =, where is the work done by the weight of the ball (see part a).
34. REASONING AND SOLUTION The conservation of energy gives Rearranging gives 1 1 mvf + mgh f = mv 0 + mgh0 ( 14.0 m/s) ( 13.0 m/s) hf h0 = = 1.4 m 9.80m/s