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Elena L [17]
3 years ago
8

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 fricti

on 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.
Physics
1 answer:
vovikov84 [41]3 years ago
3 0

Answer:

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

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}

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