1 a hulla hoop is rotted along a horizontal surface distance of 25ft. with an an

Question

1

a hulla hoop is rotted along a horizontal surface distance of 25ft. with an angular speed of 3.15 rev/sec cncounters an incline of 12dgree .neglecting any friction effects, if the diameter of the hoop is 1.33 m, how far up the incline will the hoop travel?

2

A 150 g ball is thrown vartically upward from a window 10 m above the graund at a speed of 30m/s. At the top of its trajectory, it compresses a spring of spring constant k-1.50 n/m adistance y. if the spring is 25 m above the initial starting position of the ball;

a- how high does the ball go above the ground?

b-what is the speed of the ball when it strikes the ground?

3

A muss M on an inclined plane is acted upon by a constant force F such that the block moves with a constant speed. If the incline is 23 degree and M is 2.65 kg and F = 25 N, then determine the value for the coefficient of kinetic friction.

Explanation / Answer

1. On the incline, the hula hoop experiences a deceleration of gsin12 = 2.04 m/s^2

Linear Velocity of Hoop when it starts climbing up the incline = Radius of hoop * Angular Speed (Since hoop is rolling without slipping)

v = 4.19 m/s

Eqn of motion

v'^2 = v^2 - 2as

v' = 0 (hoop stops)

a = -2.04 m/s^2

s = v^2 / 2a = 4.3 m

So, hoop travels 4.3 m on the incline plane which translates to a height of s(sin12) = 0.89 m

2. a. From starting position

Initial velocity = u = 30 m/s

Deceleration = -9.8 m/s^2

Height at which spring is attached = 25 m

So, we find final velocity of ball when it touches spring

v^2 = u^2 - 2gh

v^2 = 30^2 - 2*25*9.8

v^2 = 410

v = 20.25 m/s

Now, at this point spring is compressed to a distance y, till the velocity of ball becomes 0

This means that all the kinetic energy is converted to gravitational and spring potential energy

We write the equation

ky^2/2 + mgy = mv^2/2

y^2 + 2mgy/k - mv^2/k = 0

So, y = -mg/k +-(m^2g^2/k^2 + mv^2/k)^0.5

We neglect the negative term since y is positive

y = -mg/k + (m^2g^2/k^2 + mv^2/k)^0.5

Substituting, given values, we get

y = 5.5 m

So, total height of ball above ground = 10 + 25 + 5.5 = 40.5 m

b. Now, ball gains an initial velocity which is given to it by spring returning to initial position and gravity

ky^2/2 + mgy = mv^2/2

This gives v = 20.25 m/s

Now, ball accelerates only due to gravity for the next 35 m with an acceleration g = 9.8 m/s^2

Let speed of ball when it strikes the ground be v'

v'^2 = v^2 + 2*g*h

v'^2 = 1096

v' = 33.11 m/s

3. For block to move with a constant velocity, it's net force parallel to the incline must be zero

First, we find out normal force acting on block

N = mgcos23 + Fsin32

N = 2.65*9.8*cos23 + 25*sin32 = 37.15 N

Frictional Force acting on block = kN, where k = coefficient of friction

f = 37.15k

Now, Force parallel to plane

mgsin23 + Fcos32 = f

31.35 = 37.15k

k = 0.844

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