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

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=$9\times 10^{-2}m$

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

$T=\frac{mv^2}{r}$

Because tension is equal to centripetal force

Substitute the values

$T=\frac{0.06\times (0.63)^2}{0.47}=0.05 N$

b.$v=3.29 m/s$

$T=\frac{mv^2}{r}=\frac{0.06\times (3.29)^2}{0.09}=7.2 N$

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

$W=\frac{1}{2}m(v^2-v^2_0)$

$W=\frac{1}{2}(0.06)((3.29)^2-(0.63)^2)$

$W=0.31 J$

Answer 2

Answer:

Please see attachment

Explanation:

Please see attachment