Question

A hand pushes a 4 kg block along a table from point A to point C as shown in the figure below. The table has been prepared so that the left half of the table (from A to B) is frictionless. The right half (from B to C) has a non-zero coefficient of friction equal to u. The hand pushes the block the full distance from A to C using a constant force of 6 N. The block starts off at rest at point A and comes to a stop when it reaches point C. The distance from A to B is 0.5 m and the distance from B to C is also 0.5 m. We will consider the block by itself as our system so all forces on the block are external. The jumper has a mass of 50 kg, and the bridge's height above the river is 150 m. The rope has an unstretched length of 11 m and stretches a distance of 5 m before bringing the jumper to a stop. Determine the maximum speed of the jumper and the spring constant of the rope. In your calculation, use g = 10 N/kg. Vmax = m/s k= N/m Now calculate the maximum speed of this more massive jumper and the spring constant of the rope needed to bring the jumper to a stop after the rope stretches 5 m. (Again, use g = 10 N/kg for your calculations.) Were your predictions correct? Vmax = m/s k = N/m A hand pushes a 4 kg block along a table from point A to point C as shown in the figure below. The table has been prepared so that the left half of the table (from A to B) is frictionless. The right half (from B to C) has a non-zero coefficient of friction equal to u. The hand pushes the block the full distance from A to C using a constant force of 6 N. The block starts off at rest at point A and comes to a stop when it reaches point C. The distance from A to B is 0.5 m and the distance from B to C is also 0.5 m. We will consider the block by itself as our system so all forces on the block are external.

Answer 1

An when the jumper falls by the length equal to the length of unstretched Ghring it has maximum KE. Therefore lost in potential Energy = gain in Kinetic Energy. Im Umar = mg to. =) Imax = 2g lo = √2x10x11 = V222 m/s = max = 14.83 m/s) Ans. when he further falls by 5 m' after reaching maximum speed and have had fallen by Ilm he expenences a refarding force. Of spring and eventually comes to rest after finally falling by 16 m from the top position o ingllotd) = 1kd² d = stretch in string = 5m =) K=2mglotd) = 2x50X10X(1175) 25 1K=640N/m Aus