A solid sphere of mass 4.00kg rolls up an incline with an inclination angle of 1
ID: 1289782 • Letter: A
Question
A solid sphere of mass 4.00kg rolls up an incline with an inclination angle of 19.40. At the bottom of the incline, the center of mass of the sphere has a translational speed of 2.94 m/s.
a) What is in J, the linear kinetic energy of the sphere at the bottom of the incline?
b) What is in J, the total kinetic energy of the sphere at the bottom of the incline?
c) How far up the incline will the sphere roll? Answer in units of m.
Use 10.0 N/kg for g.
a) What is in Joules, the initial gravitational potential energy of the system relative to the ground?
b) What is in Joules, the final (the instant m2 touches the ground) gravitational potential energy of the system relative to the ground?
c) What is in Joules, the final (the instant m2 touches the ground) kinetic energy of the system?
d) Determine the speed of M1 just as M2 hits the ground. Answer in units of m/s
Use 10 N/kg for g.
a) How fast will the block be moving just before it hits the ground? Answer in units of m/s.
b) What was in m/s2 the acceleration of the mass m then?
c) What was then, in N . m, the net torque on the cylinder?
d) What was in N the magnitude of the tension on the string?
Use 10 N/kg for g.
A solid sphere of mass 4.00kg rolls up an incline with an inclination angle of 19.40½. At the bottom of the incline, the center of mass of the sphere has a translational speed of 2.94 m/s. a) What is in J, the linear kinetic energy of the sphere at the bottom of the incline? b) What is in J, the total kinetic energy of the sphere at the bottom of the incline? c) How far up the incline will the sphere roll? Answer in units of m. Use 10.0 N/kg for g. A block of mass m = 1.13 kg is suspended above the ground at a height h = 15.0 m by a spool with two arms. The spool-with- arms arrangement is a combination of a solid uniform cylinder of mass M = 3.32 kg and radius R = 0.280 m, and two rods, each of length l = 0.308 m and mass mrod = 0.410 kg. The system is released from rest. a) How fast will the block be moving just before it hits the ground? Answer in units of m/s. b) What was in m/s2 the acceleration of the mass m then? c) What was then, in N . m, the net torque on the cylinder? d) What was in N the magnitude of the tension on the string? Use 10 N/kg for g. Two masses are connected by a light string passing over a frictionless pulley as shown on Figure. The M2= 6.7 kg mass is released from rest at height h = 2.4 m . We are given that M1=2.1 kg and I, the moment of inertia of the pulley is I = 0.161 kg m2 and its radius is R = 0.290 m. a) What is in Joules, the initial gravitational potential energy of the system relative to the ground? b) What is in Joules, the final (the instant m2 touches the ground) gravitational potential energy of the system relative to the ground? c) What is in Joules, the final (the instant m2 touches the ground) kinetic energy of the system? d) Determine the speed of M1 just as M2 hits the ground. Answer in units of m/s Use 10 N/kg for g.Explanation / Answer
KE(L) = 1/2 m v^2 = 1/2 * 4 * 2.94^2 = 2 V^2 = 17.3 J linear KE
KE(R) = 1/2 I w^2 = 1/2 * 2/5 M R^2 * V^2 / R^2 = 4 V^2 / 5 where w = V / R
KE(R) = 4/5 V^2 = 6.9 J rotational KE
KE = 17.3 + 6.9 = 24.2 J total KE
M g h = 24.2
h = 24.2 / 40 = .6 m
h = L sin 19.4
L = .6 / sin 19.4 = 1.8 m
Assuming M1 is initially on the ground
PE1 = M2 g h = 6.7 * 10 * 2.4 = 161 J initial potential energy
PE2 = M1 g h = 2.1 * 10 * 2.4 = 50.4 J
KE = 161 - 51 = 110 J
KE = 1/2 (M1 + M2) * V^2 + 1/2 * .161 * V^2 / R^2 total KE
KE = 1/2 (M1 + M2 + .161 / R^2) * V^2 where omega = V / R
KE = 1/2 ( 6.7 + 2.1 + .161 / .084) V^2 = 5.4 V^2
V = (110 / 5.4)^1/2 = 4.5 m/s
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