Question

Case 1 5. Two blocks, P and Q. are stacked together on a frictionless floor as shown. The inertia of block P is greater than that of block Q. A horizontal force is applied to block P directed to the right. The two-block system moves to the right and block Q does not slide on block P. Rough surface W Frictionless floor Vio Q Free-body diagram for block P Free-body diagram for block a. In the spaces provided at right, draw and label separate free- body diagrams for blocks P and Q. (Ignore vertical forces.) b. Identify all the Newton's third law force pairs in your free- body diagrams. c. Is the magnitude of the ner force on block P greater than, less than or equal to that on block Q? Explain. Case 2 Sandpaper Frictionless floor In case 2, a piece of sandpaper of negligible inertia is placed between block P and block Q as shown. The sandpaper has a rougher surface than the blocks. The horizontal force applied to block P remains the same as in case 1. As before, block does not slide on block P. SVO PN d. Is the acceleration of the blocks in case 2 greater than, less than or equal to that in case 1? Explain. e. Is the frictional force on block Q in case 2 greater than, less than or equal to that in case 1? Explain.

Answer 1

5

a)

For P: For Q: (normal force by floor) (normal force by P) (frictional force ) (by Q on P) Tap (applied force) feo > Manormal force by Q) (frictional force by P on long (Force of gravity) (Force of gravity)

b)

n_Q and n_P

f_QP and f_PQ

c)

Since the two blocks move together, they have same acceleration. hence the net force directly depend on the mass. Since the mass of P is greater, hence net force on P is greater.

d)

the acceleration remain same since the applied force on block P has not changed.

the acceleration is determined by taking the two blocks as one. the applied force is considered on the combination. hence acceleration is given as

a = applied force/sum of mass of two blocks

Clearly, the acceleration is independent of the frictional force between the two blocks,

e)

Same.