I need a step by step solution for this problem, please. A block ml = 1.20 kg re

Question

I need a step by step solution for this problem, please.

A block ml = 1.20 kg rests on a frictionless surface. A second block m2 = 0.260 kg sits on top of the first block. The coefficient of static friction between m1 and m2 is 0.610. A horizontal force IFI A horizontal force |FI - 2.00 N is applied to the bottom block to pull it to the right as shown below. 2.00 N is applied to the bottom block to pull it to the right as shown below. (a) What is the acceleration of the blocks? (Enter the magnitude.) m/s2 (b) What is the magnitude of the static force of friction acting on the top block if it moves with the acceleration you determined in part (a)? (c) Will the top block slide on the bottom block when the bottom block is pulled to the right? To determine this, find the maximum static friction force that can possibly act on m2 magnitude direction IN --Select-- Will the block slide relative to the bottom block? Yes No

Explanation / Answer

let,

mass of the lower box, m1=1.2kg

mass of the upper box, m2=0.26kg

co-efficient of static friction m1 and m2 is, us=0.61

force applied to the lower box, F=2N

a)

use,

a=us*g

a=0.61*9.8

a=5.98 m/sec^2

b)

static frictional force,

fs=(m1+m2)*us*g

fs=(1.2+0.26)*0.61*9.8

fs=8.73 N

c)

no, the top block won't slide on the bottom block,

frictional force on m2 due to block m1 is,

F=m1*a

F=1.2*5.8

F=6.96

here,

F<fs

d)

no, motion of the top block is not realtive to the bottom block

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.