Answer:
a) = 3.94 m
b) = 3.15 m
Explanation:
Given
Mass of the block, m = 242 g
Force constant, k = 1.62 kN/m
Compression of the spring, x = 10 cm
Angle of inclination = 60°
a) if we equate the energy at the bottom of the ramp to the energy at a distance d up the ramp, we have
1/2kx² = mgh where, h = dsinΦ
1/2kx² = mgdsinΦ
1/2 * 1.62*10^3 * 0.1² = 0.242 * 9.8 * dsin 60
1/2 * 16.2 = 2.3716 * d sin 60
d sin 60 = 8.1 / 2.3716
0.866 d = 3.415
d = 3.415 / 0.866
d = 3.94 m
b) net force on the block = mgd sin 60 + µ mgd cos 60
8.1 = d[mg sin 60 + µ mg cos 60]
8.1 = d [0.242 * 9.8 * 0.866 + 0.44 * 0.242 * 9.8 * 0.5]
8.1 = d (2.05 + 0.52)
8.1 = 2.57 d
d = 8.1 / 2.57
d = 3.15 m
Answer:
Net force ( F ) = 208 N.
Speed of car ( v ) = 68.58 m/s.
Explanation:-
The radius ( r ) = 480 m.
Mass of passenger ( m ) = 65 kg.
Horizontal force ( 
) = 208 N.
Vertical force ( 
) = 637 N.
Net vertical component of force ( ) = m*g - 637 = 65 * 9.8 - 637 = 0 N.
a ) magnitude of net force .
= 208 N.
b) The speed of car.
where v is the velocity.
v² = 4704
v = 68.58 m/s.
Answer:
mass*velocity=1.5*10^4 * 15
= 22.5*10^4
Answer:
1.94601 rad/s
Explanation:
= Moment of inertia of carousel = 124 kgm²
= Angular speed of carousel = 3.5 rad/s
= Angular speed of person
r = Radius of carousel = 1.53 m
m = Mass of person = 42.3 kg
Moment of inertia of person
As the angular momentum is conserved in the system
The angular speed of the carousel after the person climbs aboard is 1.94601 rad/s
