The 50-kg boy jumps on the 5-kg skateboard with a horizontal velocity of 5 m/s. Determine the distance s the boy reaches up the inclined plane before momentarily coming to rest. Neglect the skateboard's rolling resistance.
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the question is incomplete! the angle should be given! And by using conservation of energy the K.E is 687.5 J is converted!
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Answer:
Taking the case of the boy
Mass of boy (M1)= 60kg
Initial velocity (u1) = 5m/s
Final velocity (v1)= ?? (To find)
Taking the case of skateboard
Mass of skateboard (m2) =4 kg
Initial velocity (u2) = 0 m/s. (the board is stationary)
Final velocity (v2)= ?? (To find)
Now v1 = v2 ( both are moving in the same direction and the boy is standing on the skate board so velocity will be same)
Using law of conservation of momentum
(mu)1+(mu)2=(mv)1 + (mv)2
(60*5)+(4*0)= v(M1+m2). V1 = V2 proved above
300+0=v(60+4)
300=64v
300÷64= v
V=4.68 m/s
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