Aparticle is moving in simple harmonic motion with an amplitude of 3.0 cm and a maximum velocity of 15.0 cm/s. assume the initial phase is 0. (include the sign of the value in your answer.) (a) at what position is its velocity 4.4 cm/s? cm (b) what is its velocity when its position is 1.5 cm

a)x=2.86 cm

b)V= 12.99 cm/s

Explanation:

Given that

Amplitude A= 3 cm

Maximum velocity V(max) = 15 cm/s

We know that the  speed in simple harmonic motion given as

x=Displacement from mean position

ω =Angular speed

V(max) = ωA

Now by putting the values

15 =  ω x 3

ω = 5 rad/s

a)

V= 4.4 cm/s

x=2.86 cm

b)

x= 1.5 cm

V= 12.99 cm/s

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards