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SECTION 8.4 • Changes in Mechanical Energy for Nonconservative Forces 233 ! E mech " # f k d, where d is the horizontal distance trav- eled by the skier. To find the distance the skier travels before coming to rest, we take K C " 0. With v B " 19.8 m/s and the friction force given by f k " ) k n " ) k mg, we obtain ! E mech " E C # E B " #
) k mgd 95.2 m d " v
B
2 2) k
g " (19.8 m/s) 2 2(0.210)(9.80 m/s 2 ) " " #
) k mgd (K C % U C ) # (K B % U B ) " (0 % 0) # ( 1 2 mv 2 B % 0) Example 8.9 Block–Spring Collision A block having a mass of 0.80 kg is given an initial velocity A " 1.2 m/s to the right and collides with a spring of negli- gible mass and force constant k " 50 N/m, as shown in (A) Assuming the surface to be frictionless, calculate the maximum compression of the spring after the collision. Solution Our system in this example consists of the block , it has kinetic energy and the spring is uncompressed, so the elastic potential energy stored in the spring is zero. . After the collision, when the block is at #, the spring is fully compressed; now the 1 2 mv A 2 , where the origin of coordinates x " 0 is chosen to be the equilibrium position of the spring and max is the maximum compression of the spring, which in this case happens to be x C . The total mechanical energy of the system is conserved because no nonconservative forces Because the mechanical energy of the system is con- served, the kinetic energy of the block before the collision " (B) Suppose a constant force of kinetic friction acts be- tween the block and the surface, with ) k " 0.50. If the speed of the block at the moment it collides with the spring A " 1.2 m/s, what is the maximum compression x C in the spring? Solution In this case, the mechanical energy of the system Therefore, the change in the mechanical energy of the system C is Substituting this into Equation 8.14 gives Solving the quadratic equation for x C gives x C " 0.092 m and x C " # 0.25 m. The physically meaningful root is x C " 0.092 m. The negative root does not apply to this sit- uation because the block must be to the right of the origin 25x 2 C % 3.92x C # 0.576 " 0
1 2
(50)x 2 C # 1 2
(0.80)(1.2) 2 " # 3.92x C ! E mech " E f # E i " (0 % 1 2
kx 2 C ) # ( 1 2
mv 2 A % 0) " #
f k x C ! E mech " #
f k x C " (#
3.92x C ) f k " ) k n " ) k mg " 0.50(0.80 kg)(9.80 m/s 2 ) " 3.92 N 0.15 m x max " √ m k v A " √ 0.80 kg 50 N/m (1.2 m/s) 0 % 1 2
kx 2 max " 1 2
mv 2 A % 0 K C % U s C " K A % U s A E C " E A 1 2 kx 2 " 1 2 kx 2 max E = – mv A 2 1 2 x = 0 (a) (b) (c) v C = 0 (d) x max ! " # $ E = – mv B 2 + – kx B 2 1 2 1 2 E = – mv D 2 = – mv A 2 1 2 1 2 E = – kx max 1 2 v A v B x B v D = –v A 2 Figure 8.14 (Example 8.9) A block sliding on a smooth, hori- zontal surface collides with a light spring. (a) Initially the me- chanical energy is all kinetic energy. (b) The mechanical energy is the sum of the kinetic energy of the block and the elastic po- tential energy in the spring. (c) The energy is entirely potential energy. (d) The energy is transformed back to the kinetic en- ergy of the block. The total energy of the system remains con- stant throughout the motion. |