Question

A wooden block is kept on a horizontal surface. A knife of mass 200 gm is allowed to fall on it from the height of 5 m. (From the top of wooden block). The knife penetrated 2 m inside the block. Assuming resistive force to be constant. 7a Velocity of knife just before striking wooden block in m/s is (A) 25 (B) 10 7b The resistive force offered by wooden block in N is (A) 25 (B) 10 7c Deceleration of knife inside the wood in m/s2 is (C) 5 (C) 5 (C) 5 (A) 25 (B) 10 (C) 5 Section –C (D) 20 (D) 20 (D) 20 (D) 20 (A) 25 (B) 10 7d Acceleration of knife before striking wooden block in m/s2 is

Answer 1

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Answer 2

Explanation:

A wooden block is kept on a horizontal surface. A knife of mass 200 gm is allowed to fall on it from the height of 5 m. (From the top of wooden block). The knife penetrated 2 m inside the block. Assuming resistive force to be constant. 7a Velocity of knife just before striking wooden block in m/s is (A) 25 (B) 10 7b The resistive force offered by wooden block in N is (A) 25 (B) 10 7c Deceleration of knife inside the wood in m/s2 is (C) 5 (C) 5 (C) 5 (A) 25 (B) 10 (C) 5 Section –C (D) 20 (D) 20 (D) 20 (D) 20 (A) 25 (B) 10 7d Acceleration of knife before striking wooden block in m/s2 is

•    So Mass = 200 gm = 0.2 kg
•   height h = 5 m
•  Distance d = 2 m
• a = 10 m/s^2
• Initial velocity u = 0
• So from the equation of motion we have
•                                 v^2 = u^2 + 2as
•                             v^2 = 2as
•                               v^2 = 2gh
•                              v = √2gh
•                                 = √2 x 10 x 5
•                               v = 10 m/s
• So velocity of knife before striking the wooden block is 10 m/s
• Acceleration of knife before striking the wooden block will be
•                                  a = v^2 – u^2 / 2s
•                                     = 100 – 0 / 2 x 5
•                                         = 10 m/s^2
• Now the final velocity is 0 and u = 10 m/s when knife strikes the block.
•       So v = 0
•               So v^2 = u^2 + 2as
•                  Or u^2 = - 2as
•                        100 = - 2 x a x 2
•                     100 = - 4a
•                       Or a = - 25 m/s^2
• So the deceleration of the knife inside the block is – 25 m/s^2
• Now the resistive force  
•                           F = ma
•                              = 0.2 x 25
•                         F = 5 N
• So the resistive force is 5 N

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https://brainly.in/question/16160338