First slide

Two block problem

Question

A 40 kg slab rests on a frictionless floor. A 10 kg block rests on the top of the slab as shown in figure.
The static coefficient of friction between the block and the slab is 0.60 while the kinetic coefficient is
0.40. The 10 kg block is acted upon by a horizontal force of 100 N. If g = 9.8 $m s^{- 2}$ , the resulting
acceleration of the slab will be

Moderate

A

61 $m s^{- 2}$
B

152 $m s^{- 2}$
C

147 $m s^{- 2}$
D

0.98 $m s^{- 2}$

Solution

Force of limiting friction for block $= μ_{s} m g$
$= 0.60 \times 10 \times 9.8 = 58.8 N$
If the applied force is greater than 58.8 N, then the block will move over the slab.
Kinetic friction acting on the block towards right
$= μ_{k} m g = 0.40 \times 10 \times 9.8 = 39.2 N$
This is also equal to the force of friction acting on slab towards left. This is the only force acting on slab.
So, acceleration of the sIab, $a = \frac{F}{m} = \frac{39.2}{40} = 0.98 {ms}^{- 2}$

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