Question

You push a box of mass 22.6 kg with your car up to an icy hill slope of irregular shape to a height 4.3 m. The box has a speed 12 m/s when it starts up the hill, the same time that you brake. It then rises up to the top (with no friction) to a flat area before coming to a rest when hitting a box of mass 89 kg. The heavier box then falls off a sheer cliff to the ground (with no drag). mere er de r er der en overarme time that you brake. It then rise (a) What is the speed of the heavier box as it flies off the top of the hill? (b) What is the coordinate form of the velocity of the heavier box just before hitting the ground? Û + (c) What is the vector form of the velocity of the heavier box just before hitting the ground? 2) ŷ +

Answer 1

a) Total Energy of the box at the bottom, Ebmu+PE . mv* + mgh (h:o) » į mu At the top s . (ht 4.3m) EA ? Z mu? + mghe Conservation of mechanical energy Imut, I mu* + mg ht tv. lut ght + x(12) - Lu” + 4.8 x 4.3 LU: 29.86 uz 7.72 m/s I u becomes horizontal at the top, therefore before falling off a Sheer cliff it has only horizontal Velocity vuito sui After leaving the sheer cliff its horizontal velocity remains constant and verticale velocity increases due to gravity V2: u?+2 gh 12, 0+2x9.8 X 4.3 of the box a . V = 9.18 m/s At the ground (just before hitting ) velocity ✓, uŹ + wij - 7.72 i +9.189 Scanned with ū27.724 + CamScanned