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 of 100 N. If g = 9.8 m/s², the resulting acceleration of the slab will be:(a) 0.98 m/s² (b) 1.47 m/s²(c) 1.52 m/s² (d) 6.1 m/s²

Answer 1

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Answer 2

Answer:

A) 0.98 m/s²

Explanation:

Weight of slab = 40kg ( Given)

Weight of block = 10kg ( Given)

Coefficient of friction = 0.60 ( Given)

Kinetic friction = 0.40 ( Given)

g = 9.8 m/s²

Normal reaction from 40 kg slab on 10 kg block = 10 × 9.81 = 98.1 N 

Static frictional force = 98.1 × 0.6 N is less than 100 N applied 

Since 10 kg block slides on 40 kg slab, thus the net force

= 100 N - kinetic friction 

= 100 - 98.1 × 0.4

= 61 N 

= 10 kg block will slide on 40 kg slab with 61/10 = 6.1 m/s 

Frictional force on 40 kg slab by 10 kg block = 98.1 × 0.4 = 39 N 

= 40 kg slab will move with - 39/40 m/s 

= 0.98 m/s.

Thus, the resulting acceleration of the slab will be 0.98 m/s²