Question

A 20-kg block is held at rest on the inclined slope by a peg. A 2-kg pendulum starts at rest in a horizontal position when it is released. The pendulum bob strikes the block at the bottom of its swing with a horizontal velocity of 15 m/s. The coefficient of restitution of the impact is e = 0.7, and the coefficient of kinetic friction between the block and the inclined surface is µk = 0.5. What distance does the block slide before stopping?

Answer 1

Complete Question

The diagram of this question is shown on the first uploaded image

Answer:

The distance the block slides before stopping is
`d = 0.313 \ m`

Step-by-step explanation:

The free body diagram for the diagram in the question is shown

From the diagram the angle is
`\theta = 25 ^o`

`sin \theta = (h)/(d)`

Where
`h = h_b - h_a`

So
`d sin \theta = h_b - h_a`

From the question we are told that

The mass of the block is
`m = 20 \ kg`

The mass of the pendulum is
`m_p = 2 \ kg`

The velocity of the pendulum at the bottom of swing is
`v_p = 15 m/s`

The coefficient of restitution is
`e =0.7`

The coefficient of kinetic friction is
`\mu _k = 0.5`

The velocity of the block after the impact is mathematically represented as

`v_2 f = (m_b - em_p)/(m_b + m_p) * v_2 i + ([1 + e] m_1)/(m_1 + m_2 ) v_p`

Where
`v_2 i` is the velocity of the block before collision which is 0

`= (20 - (0.7 * 2))/((2 + 20)) * 0 + ((1 + 0.7) * 2 )/(2 + 20) * 15`

Substituting value

`v_2 f = 2.310\ m/s`

According to conservation of energy principle

The energy at point a = energy at point b

So
`PE_A + KE _A = PE_B + KE_B + E_F`

Where

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

`PE_A = m_b gh_a` = 0 at the bottom

`KE _A` is the kinetic energy at A which is mathematically represented as

`K_A = (1)/(2) m_b * v_2f^2`

`PE_B` is the potential energy at B which is mathematically represented as

`PE_B = m_b gh`

From the diagram
`h = h_b -h_a`

`PE_B = m_b g(h_b - h_a)`

`KE _B` is the kinetic energy at B which is 0 (at the top )

Where is
`E_F` is the workdone against velocity which from the diagram is

`\mu_k m_b g cos 25 *d`

So

`(1)/(2) m_b v_2 f^2 = m_b g h_b + \mu_k m_b g cos \25 * d`

Substituting values

`(1)/(2) * 20 * 2.310^2 = 20 * 9.8 * d sin(25) + 0.5* 20 * 9.8 * cos 25 * d`

So

`d = 0.313 \ m`