a block of mass m  5.00 kg is
pulled along a horizontal frictionless floor by a cord
that exerts a force of magnitude F  12.0 N at an angle
u  25.0°. (a) What is the magnitude of the
block’s acceleration? (b) The force magnitude F is
slowly increased. What is its value just before the
block is lifted (completely) off the floor? (c) What is
the magnitude of the block’s acceleration just before
it is lifted (completely) off the floor?

Question

a block of mass m  5.00 kg is
pulled along a horizontal frictionless floor by a cord
that exerts a force of magnitude F  12.0 N at an angle
u  25.0°. (a) What is the magnitude of the
block’s acceleration? (b) The force magnitude F is
slowly increased. What is its value just before the
block is lifted (completely) off the floor? (c) What is
the magnitude of the block’s acceleration just before
it is lifted (completely) off the floor?

anonymous · Answer

Resolve 12 N force into its components. Horizontal force will be  12cos25. so answer to first part is a=F/m = 12cos25/5

anonymous · Answer

now vertical component of force is 12sin25. in order to life upwards it must overcome the weight of the body Let the force should be increased to F. then F sin25=5x9.8 and hence F- 49/sin 25. now for third part acceleration can be calculated as a=F/m