Question

A 3 kg block has a speed of 4m/s and 8m/s at A and B respectively. If the distnce between A and B along the curve is 12m, the calculate the magnitude of frictional force acting on the block. Assuming the same friction, how far away from B the block will stop?

Answer 1

нєу тнєяє ιѕ αиѕωєя !!!
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ι нσρє уσυ нєℓρ !!
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As per given in the question:
Distance:                                 s = 12 m
Mass:                                      m = 3 kg
Formula: third equation of motion:
                                              
                                                                      v2 = u2 + 2aswhere a = accelerationputting the values in above equation:42 = 82 + 2×a×1216 = 64 + 24aa = −4824 = −2 ms2          ......(1)/So the friction force acting on the body:Fr = m×a = 3×2 = 6 N [Dir. of friction force is opposite to the dir. of motion of the body]
now we will calculate the distance traveled after point A:
                          v = 0
                         u = 4
                         a = -2
                        s = ?
third eq. of motion:
                                                                      v2 = u2 + 2as02 = 42 + 2×−2×ss = 164s = 4 m

Answer 2

As per the law of conservation of energy the mechanical energy remains conserved.

so we will calculate the mechanical energy at point A and B

point A :- mgh + ½mv²

=3(10)6 + ½(3)(4)²

= 204 J

point B :- mgh + ½mv²

= 3(10)2 + ½(3)(8)²

= 156J

here we see a different in mechanical energy

so we know the work done by frictional force

i.e. 204J - 156J

= 48J

now the distance between A and B is 12m and workdone by friction is 48J

so, W = Fs

W/s = F

F = 48/12

F = 4N

therefore the magnitude of friction is 4N

hope it helps u!

have a nice day