A horizontal block-spring system with the block on a frictionlesssurface has tot

Question

A horizontal block-spring system with the block on a frictionlesssurface has total mechanical energy E = 43.8 J and a maximum displacement from equilibrium of0.205 m. (a) What is the spring constant?
                                 N/m

(b) What is the kinetic energy of the system at the equilibriumpoint?
                                    J

(c) If the maximum speed of the block is 3.45 m/s, what is itsmass?
                                      kg

(d) What is the speed of the block when its displacement is 0.160m?
                                       m/s

(e) Find the kinetic energy of the block at x = 0.160m.
                                          J

(f) Find the potential energy stored in the spring when x= 0.160 m.
                                          J

(g) Suppose the same system is released from rest at x =0.205 m on a rough surface so that itloses 13.2 J by the time it reaches itsfirst turning point (after passing equilibrium at x = 0).What is its position at that instant?
                                     m (a) What is the spring constant?
                                 N/m

(b) What is the kinetic energy of the system at the equilibriumpoint?
                                    J

(c) If the maximum speed of the block is 3.45 m/s, what is itsmass?
                                      kg

(d) What is the speed of the block when its displacement is 0.160m?
                                       m/s

(e) Find the kinetic energy of the block at x = 0.160m.
                                          J

(f) Find the potential energy stored in the spring when x= 0.160 m.
                                          J

(g) Suppose the same system is released from rest at x =0.205 m on a rough surface so that itloses 13.2 J by the time it reaches itsfirst turning point (after passing equilibrium at x = 0).What is its position at that instant?
                                     m

Explanation / Answer

(a) What is the spring constant?                                  N/m Energy = KineticE + PotentialE = 0.5*m*v2 +0.5*k*x2. At x = x_max, the object is momentarily stationary. And all thekinetic energy is converted to potential (KE = 0) Energy = PotentialE = 0.5*k*x2. = 43.8 J k = 43.8J/(0.5*(0.205m)2) = 2084 N/m (b) What is the kinetic energy of the system at the equilibriumpoint?                                     J At the equilibrium point, x = 0; so E = KE = 43.8 J   (nopotential energy so all energy is kinetic) (c) If the maximum speed of the block is 3.45 m/s, what is itsmass?                                       kg At max speed, all energy is kinetic energy so KE = 0.5*m*vmax2 = 43.8 J m = 43.8 J/(0.5*(3.45m/s)2) = 7.36 kg (d) What is the speed of the block when its displacement is 0.160m?                                        m/s E = 0.5*m*v2+0.5*k*x2 v2 = (E - 0.5*k*x2)/(0.5*m) v2 = (43.8 -0.5*2084*0.1602)/(0.5*7.36) v2 = 4.65 v = 2.16 m/s (e) Find the kinetic energy of the block at x = 0.160m.                                           J KE = 0.5*m*v2 = 0.5*7.36*(2.162) = 17.2 J (f) Find the potential energy stored in the spring when x= 0.160 m.                                           J PE = E - KE = 43.8 - 17.17 = 26.6 J (g) Suppose the same system is released from rest at x =0.205 m on a rough surface so that itloses 13.2 J by the time it reaches itsfirst turning point (after passing equilibrium at x = 0).What is its position at that instant?

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.