First slide

Two block problem

Question

A 40 kg slab rests on a frictionless floor. A l0 kg block rests on top of the slab. The static coefficient of friction between the block and the slab is 0.60 while the kinetic coefficient of friction is 0.40. The 10 kg block is acted upon by a horizontal force of 100 N. The resulting acceleration of the slab will be

Moderate

A

1.5 ${ms}^{- 2}$
B

6.0 ${ms}^{- 2}$
C

10 ${ms}^{- 2}$
D

1.0 ${ms}^{- 2}$

Solution

The maximum force for which there will be no relative motion between the blocks is given by $F_{m a x =} μ m g [1 + \frac{m}{M}] = 0.6 \times 10 \times 10 [1 + \frac{10}{40}] N = 75 N S i n c e t h e a p p l i e d f o r c e (100 N) i s g r e a t e r t h a n F_{m a x}, t h e 10 k g b l o c k w i l l s l i p o v e r 40 k g b l o c k . T h e r f o r e a c c e l e r a t i o n o f 10 k g b l o c k = \frac{100 - 0.4 \times 10 \times 10}{10} = 6 m / s^{2}$

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