Please Answer STEP BY STEP!! Mass m sits on a frictionless, horizontal table. As
ID: 2164467 • Letter: P
Question
Please Answer STEP BY STEP!!Mass m sits on a frictionless, horizontal table. Assume:
- All vertical forces acting on m (such as gravity and the normal force) sum to zero.
- All applied forces act parallel to the table
When a force of magnitude F is applied to mass m, it accelerate at magnitude a = 7.62 m/s2.
NOTE: The axes lie along the table.
a) Force F1 has magnitude F and points in the positive x-direction; force F2 also has magnitude F and points in the positive y-direction. If both of these forces act on mass m at the same time, what will the magnitude of its acceleration be now?
b) Suppose force F3 has magnitude 2F and points at 45o to the x-axis. If forces F1 and F3 act on mass m at the same time, what will the magnitude of its acceleration be now?
Explanation / Answer
To solve this problem you must invoke some physical insight to get a feeling of what to do. Initially, the only energy in the block-wedge system is stored in the block, but afterwards the block moves with speed v1 and the wedge moves with speed v2. This is assuming that the wedge does not sit there stationary and moves when the block slides down By the conservation of energy, the total mechanical energy is conserved. Eo = Ef K_block + K_ramp + U_block = K_block + K_ramp + U_block mgh = (1/2)mv1 ^2 + (1/2)Mv2 ^2 The only external force acting on the block is gravity but that is in the y direction. The system's x- momentum is conserved though. Po = Pf 0 = mv1 + Mv2 v2 = - (m1/M)v1 = -(0.8/6)*(2.4) = - 0.32 m/s -0.32 m/s is the speed of the wedge right after the block reaches the horizontal surface. There was no initial velocity for either the block or the wedge Using this information, you can find the height h of the wedge just use the formula h = [ (1/2)mv1^2 + (1/2)Mv^2 ]/mg h = [ (1/2)(0.8)(2.4)^2 + (1/2)(6)(-0.32)^2 ]/ (0.8)*(9.8) =