First slide

Kinetic friction

Question

A 40 kg slab rests on frictionless floor as shown in fig. A 10 kg block rests on the top of the slab.
The static coefficient of friction between the block and slab is 0.60 while the kinetic friction is 0.40. The 10 kg block is acted upon by a horizontal force 100 N. If g = 9.8 m / s², the resulting acceleration of the slab will be

Moderate

A

0.98 m/s²
B

1.47 m/s²
C

1.52 m/s²
D

6.1 m/s²

Solution

The frictional force between 10 kg block and slab of 40 kg is given by $f = μmg = 0 \cdot 4 \times 10 \times 9 \cdot 8 N .$
The acceleration of the slab of 40 kg is $a = \frac{0 \cdot 4 \times 10 \times 9 \cdot 8}{40} = 0 \cdot 98 m / s^{2} .$

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