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Orlov [11]
3 years ago
13

Five wheels are connected as shown in the figure. Find the velocity of the block “Q”, if it is known that: RA= 5 [m], RB= 10 [m]

, RD= 6 [m], RE=12 [m]. ​

Physics
1 answer:
Tanzania [10]3 years ago
6 0

Answer:

-5 m/s

Explanation:

The linear velocity of B is equal and opposite the linear velocity of E.

vB = -vE

vB = -ωE rE

10 m/s = -ωE (12 m)

ωE = -0.833 rad/s

The angular velocity of E is the same as the angular velocity of D.

ωE = ωD

ωD = -0.833 rad/s

The linear velocity of Q is the same as the linear velocity of D.

vQ = vD

vQ = ωD rD

vQ = (-0.833 rad/s) (6 m)

vQ = -5 m/s

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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
gregori [183]

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

Explanation:

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

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

         sin \theta  = \frac{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 = \frac{m_b - em_p}{m_b + m_p}  * v_2 i + \frac{[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

                  = \frac{20 - (0.7 * 2)}{(2 + 20)} * 0 + \frac{(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 = \frac{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

   \frac{1}{2} m_b v_2 f^2  = m_b g h_b + \mu_k m_b g cos \25 * d

Substituting values

   \frac{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

       

   

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Answer:

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