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brilliants [131]
3 years ago
7

A 40 kg slab rests on a frictionless floor. A 10 kg block rests on top of the slab. The coefficient of static friction between t

he block and the slab is 0.60, whereas the kinetic coefficient is 0.40. The 10 kg block is acted on by a horizontal force of 100 N. Calculate the resulting accelerations of (a) the block and (b) the slab.

Physics
1 answer:
Wewaii [24]3 years ago
3 0

Answer:

a) a = 6.1 m/s^2

b)  a = 0.98m/s^2

Explanation:

Mass of slab = 40kg

Mass of block = 10kg

Coefficient of static friction (Us)  = 0.60

Kinetic coefficient (UK)  = 0.40

Horizontal force = 100N

The normal reaction from 40kg slab on 10 kg block = 10*9.81

= 98.1N

Static frictional force = Us*R

= 98.1*0.6

= 58.86N

This is less than the force applied

If 10 kg block will slide on the 40 kg slab,  net force = 100 - kinetic force

Kinetic force (Uk*R) = 0.4*98.1

= 39.28N

= 39N

Net force = 100 -39

= 61N

Recall that F = ma

For 10 kg block

a = F/m

a = 61/10

a = 6.1m/s^2

b) Frictional force on 40 kg slab by 10 kg = 98.1*0.4

= 39.24

= 39N

F = ma

a = F/m

For 40kg slab

a = 39/40

a = 0.98m/s^2

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