Question

QuestioN NO. 2
A wooden block is kept on a horizontal surface. A knife of mass 200 g is allowed to fall from 15 m height from
the top surface of the wooden block. The knife penetrated 5 m inside the block. Assumes resistance force to be
constant. Mark the correct options

a. Just before striking wooden block velocity of knife is 17.32 m/s.

b. The resistive force offered by block is 4 N.

c. Deceleration of knife inside the block is 30m/s-

d The resistive force offered is 8 N.
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Answer 1

Answer:

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D) the resistive force offered is 8N

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Answer 2

Option B & D are incorrect.

Explanation:

Mass = 200 gm = 0.2 kg

h = 15 m

d = 5 m

From the third law of motion

$v^{2} = u^{2} + 2as$

Here a = g = 9.81 $\frac{m}{s^{2} }$

Initial velocity (u) = 0

Distance s = h = 15 m

So the final velocity

$v = \sqrt{2gh}$

$v = \sqrt{2 (9.81)(15)}$

v = 17.32 $\frac{m}{s}$

This is the velocity of knife Just before striking wooden block.

Thus option (a) is correct.

When the knife strikes the block the final velocity of knife becomes zero. i.e.

v = 0 & u = 17.32 $\frac{m}{s}$ in that case.

$v^{2} = u^{2} + 2ad$

Since v = 0 so

$u^{2} = -2ad$

$a = \frac{-u^{2} }{2d}$

Put the values of u & d in above equation we get

$a = - \frac{(17.32)^{2} }{(2)(5)}$

a = - 30  $\frac{m}{s^{2} }$

This is the deceleration of the knife inside the block.

Option (C) is also correct.

The resistive force applied by the block is

F =  ma

F =  (0.2) × 30

F = 6 N

This is the resistive force applied by the block.

Thus option B & D are incorrect.