A solid sphere rolls without slipping from rest down a plane 4.00 m long incline

Question

A solid sphere rolls without slipping from rest down a plane 4.00 m long inclined at an angle of 42.0 degree with respect to the horizontal as shown in the figure below. Determine the speed at the bottom of the incline. A) 8.85 m/s B) 6.13 m/s C) 7.25 m/s D) 7.48 m/s E) 5.13 m/s Two masses m_1 and m_2 are connected by a string which goes over an ideal pulley as shown in the figure above. Block m_1 has a mass of 3.00 kg and can slide along a rough plane inclined 30.0 degree to the horizontal. The coefficient of kinetic friction between block m_1 and the plane is 0.400. Block m_2 has a mass of 2.77 kg. What is the acceleration of each mass? A) 0.040 m/s^2 B) 0.392 m/s^2 C) 4.36 m/s^2 D) 0.728 m/s^2 E) 0.0742 m/s^2 The internal energy of an ideal gas depends on A) its temperature, pressure, and volume. B) its temperature. C) its temperature and pressure. D) its pressure. E) its volume. A child is riding a merry-go-round, which has an instantanteous angular speed of 1.25 rad/s and an angular acceleration of 0.745 rad/s^2. The child is standing 4.65 m from the center of the merry-go-round. What is the magnitude of the net acceleration of the child? A) 7.27 m/s^2 B) 8.05 m/s^2 C) 3.46 m/s^2 D) 2.58 m/s^2 E) 4.10 m/s^2

Explanation / Answer

Given:

length of incline = s= 4m

angle of incline = 42

so height of the incline h= s sin 42=4 x sin 42

if m= mass of the sphere, then loss in p.e. = mgh= mg x 4 x sin 42

gain in kinetic energy = gain in linear energy + gain in rotational energy

= 1/2 x m x v2  + 1/2 x (0.4 mr2) w2 where 0.4 mr2 is the moment of inertia of sphere

= 1/2 x m x v2  + 1/2 x (0.4 mr2) (v/r)2

=0.7 m x v2  

as energy is conserved. loss in potential energy= gain in kinetic energy

so, mgx4xsin42 =0.7 m x v2  

so, v= 6.13 m/s

so option b is correct

all the best in the course work.

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