First slide

Friction on horizontal surface

Question

If 100 N force is applied to 10kg block as shown in diagram, then acceleration produced for slab :

Moderate

A

1.65 m/s²
B

0.98 m/s²
C

1.2 m/s²
D

0.25 m/s²

Solution

Let friction force between block and slab be f and both block move together with acceleration a.
$a = \frac{100}{40 + 10} = \frac{f}{40} \Rightarrow f = 80 N$
As $f_{max} = μ_{s} (10 g) = 0 .6 \times (10 \times 9 .8) = 58 .8 N$ which is less than 80 N, so there is slipping between block and slab. Therefore acceleration produced in slab
$= \frac{μ_{k} (10 g)}{40} = 0 .98 m / s^{2}$

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