Question

A small block with a mass of 0.0600 kg is attached to a cord passing through a hole in a frictionless, horizontal surface (Figure 1) . The block is originally revolving at a distance of 0.47 m from the hole with a speed of 0.63 m/s . The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 9.0×10?2 m . At this new distance, the speed of the block is 3.29 m/s .

A small block with a mass of 0.0600 kg&n

Part A.) What is the tension in the cord in the original situation when the block has speed v0 = 0.63 m/s ?(T=?)

Part B.) What is the tension in the cord in the final situation when the block has speed v1 = 3.29 m/s ?(T=?)

Part C.) How much work was done by the person who pulled on the cord? (W=?)

I really need to understand this :o

Answer 1

Please see attachment

Step-by-step explanation:

Please see attachment

Answer 2

Answer with Explanation:

We are given that mass of block=0.0600 kg

Initial speed of block=0.63 m/s

Distance of block from the hole when the block is revolved=0.47 m

Final speed=3.29 m/s

Distance of block from the hole when the block is revolved=
`9* 10^(-2)m`

a.We have to find the tension in the cord in the original situation when the block has speed =
`v_0=0.63 m/s`

`T=(mv^2)/(r)`

Because tension is equal to centripetal force

Substitute the values

`T=(0.06* (0.63)^2)/(0.47)=0.05 N`

b.
`v=3.29 m/s`

`T=(mv^2)/(r)=(0.06* (3.29)^2)/(0.09)=7.2 N`

c.Work don=Final K.E-Initial K.E

`W=(1)/(2)m(v^2-v^2_0)`

`W=(1)/(2)(0.06)((3.29)^2-(0.63)^2)`

`W=0.31 J`