A 2kg of block of wood rests on a long table top. A 5 g bullet moving horizontally with a speed of 150 m/s is shot into the block and sticks to it. The block then slides 2.7 m along the table top and comes to a stop. The force of friction between the block and the table is:
Answers
The force of friction between the block and the table is F = 0.052 N
Explanation:
Given data:
Mass of block "m" = 2 Kg
Mass of bullet "m" = 5 g
Distance covered by block = 2.7 m
Solution:
Velocity of block just after collision will be
v = 5×10^−3×150 / 2+5×10^−3
v = 750 x 10^-3 / 7 x 10^-3
v = 0.374 m /s
Let, F be the force of friction. Then, work done against friction = initial kinetic energy.
F × 2.7 = 1 / 2 × 2.005×(0.374)^2
F = 0.052 N
Hence the force of friction between the block and the table is F = 0.052 N
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A particle has initial velocity (2i+3j) and acceleration (0.3i+0.2j) the magnitude of the velocity after 10 second will be ?
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Answer:
0.052 N
Explanation:
Mass of block (m) = 2 Kg
Mass of bullet (m1) = 5 g = 0.005 kg
Speed of bullet (v1) = 150 m/s
Distance slide by block (s) = 2.7m
To Find :- frictional force (f) between block and table
Solution :- Let the Speed of sliding of block on table top be v2
Since we know , By linear momentum i.e,
m1v1=m2v2
⟹ (0.005) × 150 = (2.005) × v2
∴ v2 = 0.374 m/s
Now , By work energy theorem we get ,
f×s=21(m+m1)(v2)2
⟹ f×(2.7)=21(2.005)(0.374)2
∴ f=0.052N
Hence , Option A (0.052N) is correct.