Starting at x = 0 m, a 1.5-kg object is moving at 6.0 m/s in the positive x direction. During its motion, the object encounters a variable force as shown in the graph that is parallel with the x-axis. Determine the final speed of the object when it is at x = 24 m. 9 8 7 6 Force (N) 01 5 0 -1 -2 -3 -4 -5 O 7.6 m/s 4.1 m/s ○ O 6.4 m/s O 5.7 m/s O 3.6 m/s 0 2 4 (O 8 10 12 14 16 Displacement (m) 18 20 22 24

A force F = (1.70 N) i + (5.70 N)j+(7.80 N) k acts on a 5.50 kg mobile object that moves from an initial position of d = (7.50 m) i+(9.10 m)j+ (8.80 m) k to a final position of d f = (4.90 m) î + (6.00 m)j + (9.40 m) k in 2.90 s. Find (a) the work done on the %3D %3D %3D object by the force in the 2.90 s interval, (b) the average power due to the force during that interval, and (c) the angle between vectors d, and d .

A force F = (4.40 N) î + (3.00 N)ĵ + (1.80 N) k acts on a 5.40 kg mobile object that moves from an initial position of á ; = (4.90 m) î + (3.50 m)ĵ + (5.90 m) k to a final position of d; = (3.20 m)î + (6.60 m)ĵ + (5.10 m) k in 2.20 s. Find (a) the work done on the object by the force in the 2.20 s interval, (b) the average power due to the force during that interval, and (c) the angle between vectors d ; and d f. (a) Number Units (b) Number Units (c) Number i Units >

A force É = (5.90 N )î + (5.40 N )f + (6.40 N )k acts on a 1.20 kg mobile object that moves from an initial position of (4.20 m )î + (2.10 m )ŷ + (7.40 m )k to a final position of á f = (5.90 m )î + (8.20 m )ŷ + (5.50 m )k in 5.70 s. i Find (a) the work done on the object by the force in the 5.70s interval, (b) the average power due to the force during that interval, and (c) the angle between vectors d ; and d f. i (a) Number i Units (b) Number i Units (c) Number i Units

Figure P7.44 45. Review. Two constant forces act on an object of mass m = QIC 5.00 kg moving in the xy plane as shown in Figure P7.45. Force F, is 25.0 N at 35.0°, and force F, is 42.0 N at 150°. At time t = 0, the object is at the origin and has velocity (4.00î + 2.50j) m/s. (a) Express the two forces in unit-vector notation. Use unit-vector notation for your other answers. (b) Find the total force exerted on the object. (c) Find the object's acceleration. Now, considering the instant t = 3.00 s, find (d) velocity, (e) its position, its kinetic energy the object's (f) from mv, and (g) its F2 F kinetic from 150° energy mu + EF · A. (h) What conclusion can you draw by comparing the answers to parts (f) and (g)? 35.0° m Figure P7.45

After being hit, a 950 g rock with an initial speed of 455 cm/s slides along a horizontal road against a friction force of 1.91 N. a) How long will it take before it stops? b) What is the coefficient of kinetic friction between the surfaces? c) How much kinetic energy did the rock possess after being hit?

Figure 11 8. A skier (68 kg) starts from rest but then begins to move downhill with a net force of 92 N for 8.2 s. The hill levels out for 3.5 s. On this part of the hill, the net force on the skier is 22 N [backwards]. ™ (a) Calculate the speed of the skier after 8.2 s. (b) Calculate the speed of the skier at the end of the section where the hill levels out. (c) Calculate the total distance travelled by the skier before coming to rest.