Question

A 40 kg slab rests on a frictionless floor. A 10 kg block rests on top of the slab (Fig. 6-58). The coefficient of static friction µstat beltween the block and the slab is 0.70, whereas their kinetic friction coefficient µkin is 0.40. The 10 kg block is pulled by a horizontal force with a magnitude of 100 N.

Answer 1

Answer:

$\text { The "resulting action" on the slab is } 0.98 \mathrm{m} / \mathrm{s}^{2}$

Explanation:

Normal reaction from 40 kg slab on 10 kg block

M × g  = 10 × 9.8 = 98 N  

Static frictional force = 98 × 0.7 N

Static frictional force = 68.6 N is less than 100 N applied  

10 kg block will slide on 40 kg slab and net force on it  

= 100 N - kinetic friction  

$=100-(98 \times 0.4)\left(\mu_{\text {kinetic }}=0.4\right)$

= 100 - 39.2

= 60.8 N

$10 \mathrm{kg} \text { block will slide on } 40 \mathrm{kg} \text { slab with, } \frac{\mathrm{Net} \text { force }}{\text { mass }}$

$10 \mathrm{kg} \text { block will slide on } 40 \mathrm{kg} \text { slab with }=\frac{60.8}{10}$

$10 \mathrm{kg} \text { block will slide on } 40 \mathrm{kg} \text { slab with }=6.08 \mathrm{m} / \mathrm{s}^{2}$

$\text { Frictional force on 40 kg slab by 10 kg block, normal reaction \times \mu_{kinetic } }$

Frictional force on 40 kg slab by 10 kg block = 98 × 0.4  

Frictional force on 40 kg slab by 10 kg block = 39.2 N  

$40 \mathrm{kg} \text { slab will move with } \frac{\text { frictional force }}{\text { mass }}$

$40 \mathrm{kg} \text { slab will move with }=\frac{39.2}{40}$

40 kg slab will move with = $0.98 \mathrm{m} / \mathrm{s}^{2}$

$\text { The "resulting action" on the slab is } 0.98 \mathrm{m} / \mathrm{s}^{2}$