Question

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 the 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.

Answer 1

a) a = 6.1 m/s^2

b) a = 0.98m/s^2

Step-by-step 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