# A frictionless vertical post is inserted into a bowling ball with a hole drilled through its center. The ball has a mass of 10.5 kg. A rope is attached to a ring that has been affixed to the...

A frictionless vertical post is inserted into a bowling ball with a hole drilled

through its center. The ball has a mass of 10.5 kg. A rope is attached to a

ring that has been affixed to the ball. Kurt pulls on the rope at an angle of

(theta)= 24.5° with respect to the vertical with a constant force of

magnitude . F=134N

(a) How much work does Kurt do on the bowling ball by raising it through a vertical distance of 1.08 m at a

constant velocity?

(b) How much work does the Earth do on the ball while it is being raised through a distance of 1.08 m?

(c) Kurt is lazy and would like to be able to raise the ball through the same vertical distance with a minimum

effort. What two changes in his actions would you recommend to her?

(d) What is the net work done on the ball by the two forces acting on it?

(e) If Kurt raises the mass in a time of 0.56 s what is his horse power?

### 1 Answer | Add Yours

The figure is attached below.

b)

Since the gravitational forces are conservative, the work done by them on a mass does not depend on the path taken by the mass, but only on the starting and ending points of the trajectory. Thus the work done by the gravitational force (by the Earth) on the ball that raises for a height `h` is just

`W_g =-G*h =-m*g*h =-10.5*9.81*1.08 =-111.25 J`

The minus sign comes from the fact that the weight of the ball is downwards while the upper motion of the ball is upwards.

a)

Kurt is raising the ball through a height `h` and at the same time is displacing the ball on the horizontal through a distance `d` .

The length of the rope `L` can be found from

`h = L-L*cos(alpha) rArr L =h/(1-cos(alpha)) =1.08/(1-cos(24.5)) =12 m`

The horizontal distance `d` is

`d =L*sin(alpha) =12*sin(24.5) =4.98 m`

The force exerted by Kurt can be decomposed along vertical and horizontal axes:

`F_x =F*sin(alpha) =134*sin(24.5) =55.57 N`

`F_y = F*cos(alpha) =134*cos(24.5) =121.93 N`

The work done by Kurt is

`W =F_x*d -F_y*h =55.57*4.98-121.93*1.08=145.05 J`

Again the minus sign in front of the second term comes from the fact that the vertical force `Fy` exerted by Kurt is opposing the displacement upwards on the vertical the height `h` of the ball.

c)

The do the minimum effort in raising the ball Kurt needs to pull directly upwards on the ball and change the magnitude of the pulling force to `F=m*g =10.5*9.81 =103 N`

d)

The net work done on the ball by the gravitational forces and by Kurt is

`W_("tot") =Wg+W =-111.25+145.05 =33.8 J`

e)

`P =W/t =145.05/0.56 =259.02 W =259.02/745.7 =0.347 HP`

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