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[tex]The rear side of a truck is open and a box of mass 20 Kg is placed on a truck 4 metre away from the open end.___________❤ü= 0.15g=10m/s²___________❤The truck starts from rest with an acceleration of 2m/s² on a straight road.At what distance the box will fall off the truck from the starting point ? |
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Answer» ESTION ✪✪||The rear side of a truck is OPEN and a box of mass 20 Kg is PLACED on a truck 4 metre AWAY from the open end.ü= 0.15g=10m/s²The truck starts from rest with an acceleration of 2m/s² on a straight road. At what distance the box will fall off the truck from the starting point ?|| ✰✰ ANSWER ✰✰ ||Given That :- → m = 40 kg→ a = 2 m/s²→ μ = 0.15→ d = 4m. Here, net force acting on the box is given by,Net Force = Force DUE to speed of truck - Frictional force→ Fnet = Fx - Ff→ Fnet = ma - μmg→ Fnet = 40×2 - 0.15×40×10→ Fnet = 80 - 60→ Fnet = 20 NNet acceleration = Fnet/m→ a' = 20/40 = 0.5 m/s²So ,TIME required for box to fall off :- s = ut + 1/2 a't²→ 4 = 0×t + (1/2)×0.5×t²→ 4 = 0.25t²→ 4 = (1/4)t²→ t² = 4 * 4 → t = 4 seconds. So, After 4 seconds The box will Fall of The Truck and ,,Distance travelled by truck in 4 s :- → s' = ut + 1/2 at²→ s' = 0×4 + (1/2)×2×4²→ s' = 0 + 4²→ s' = 16m . Hence, The Truck will travel 16m before box falling off. |
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