A small block with mass 0.200 kg is released from rest at the top of a frictionl

Question

Explanation / Answer

mass of the block released from rest at the top of a frictionless incline, m = 0.2 kg

distance travels by the block, x = 0.667 m

time taken, t = 2 sec

if 0.2 kg mass is replaced by a 0.4 kg released from rest at the top of the incline, then distance travels by blockwjich is given as ::

using equation of motion 2,

x = v0t + 1/2 a t2   Where, v0 = initial velocity = 0 m/s, t = 2 sec, x = 0.76 m

inserting all these values in above eq.

(0.667 m) = 0 + (0.5) a (2 sec)2

a = (0.667) / (2 s2)

a = 0.335 m/s2

again, using equationof motion1,

vf = v0 + a t                                

vf = 0 + (0.335 m/s2) (2 sec)

vf = 0.667 m/s

now, average velocity is given as :

vavg = (vf + v0) / 2                            

inserting the values in eq.3

vavg = [(0.667) + (0 m/s)] / 2

distance by the block moves in 2 s which given as :

D = vavg t                              

D = 0.76 m

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