Question

Help!!

A 0.368 kg bead can slide on a curved wire as seen in the figure.
Assume h1 = 4.20 m and h2 = 2.14 m. If the wire is frictionless and the bead is released with an initial speed of 1.96 m/s in the forward direction then find the speed of the bead at B.
If the wire is frictionless, find the speed of the bead at C.

Answer 1

The velocity at B is
`v = 12.6 \ m/s`

The velocity at C is
`v_c =10.80 \ m/s`

Step-by-step explanation:

From the question we are told that

The mass of the bead is
`m_b = 0.368 \ kg`

The first height is
`h_1 = 4.20 \ m`

The second height is
`h_2 = 2.14 \ m`

The initial speed of the bead is
`u = 1.96 \ m/s`

According to the law of energy conservation

`KE _ A + PE_A = KE_B + PE_B`

Now
`KE _ i` is the kinetic energy at A and it is mathematically represented as

`KE_A = (1)/(2) mu^2`

`PE_i` is the potential energy at A which is mathematically represented as

`PE_A = mg h_1`

`KE_f` is the kinetic energy at B which is mathematically represented as

`KE_B = (1)/(2) mv^2`

Where v is the speed of the bead at B

`PE_B` is the potential energy at B which equal to 0 because height is 0 at B

So

`(1)/(2) mu^2 + mg h_1 = (1)/(2) mv^2`

Making v the subject

`v = √(u ^2 + 2gh_1)`

substituting values

`v = √(1.96 ^2 + 2* 9.8 * 4.20)`

`v = 12.6 \ m/s`

According to the law of energy conservation

`KE _ B + PE_B = KE_C + PE_C`

So

`KE_B = (1)/(2) m v^2`

`PE_B = 0`

`KE_C` is the kinetic energy at c which is mathematically represented as

`KE_C = (1)/(2) * m v_c^2`

`PE_C` is the potential energy at C which is mathematically represented as

`PE_C = mg h_2`

So

`(1)/(2) m v^2 = (1)/(2) * m v_c^2 + mg h_2`

`v^2 = v_c^2 + 2g h_2`

making
`v_c` the subject

`v_c = √(12.6^2 - 2* 9.8 * 2.14)`

`v_c =10.80 \ m/s`