Question

A 2.5 kg block is initially at rest on a horizontal surface.A horizontal force F of magnitude 6.0N and a vertical force P are then applied to the block. The coefficients of friction for the block and surface are μs =0.40 and μk=0.25.Determine the magnitude of the frictional force acting on the block if tha mgnitude of P is i)8.0N ii)10 N and iii) 12 N

Answer 1

6》□ fmeusmax = 0.40×25 = 10
F<=fmeumax
so a in x = 0
a in y will not change .
frictional force in x direction will be remain same in all cases. i.e 6
in case 1
F in y <= fmeumax
so overall force = 8 +6=14
in case 2
F = fmeumax
so a = 0
so overall force = 10 +6 =16
in case 3
F>fmeumax
so a cannot be zero
fk = 0.25×2.5
= 0.625
ma =12- fk
=12-0.625
=11.375 +6 =17.375

Answer 2

Answer:

A 2.5 kg block is initially at rest on a horizontal surface. A horizontal force of magnitude 6.0 N and a

vertical force are then applied to the block (Fig. 6-17).The coefficients of friction for the block and

surface are µs = 0.40 and µk = 0.25. Determine the magnitude of the frictional force acting on the block

if the magnitude of P is (a) 8.0 N, (b) 10 N, and (c) 12 N.

Explanation:

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