Question

A 40 kg slab rests on a friction less floor.a 10 kg block rests on top of the slab.the coefficient kinetic friction between the block and the slab is 0.40.a horizontal force of 100N is applied on the 10 kg block.find the resulting acceleration of the slab.

Answer 1

m*g = 10kg * 9.8N/kg = 98 N. = Wt. of the block.

Fk = u*mg = 0.40 * 98 = 39.2 N. = Force
of kinetic friction.

a = Fk/Ms = 39.2/40 = 0.98 m/s^2.

Answer 2

Normal reaction from 40 kg slab on 10 kg block = 10 * 9.81 = 98.1 N
Static frictional force = 98.1 * 0.6 N is less than 100 N applied
=> 10 kg blck will slide on 40 kg slab and net force on it
= 100 N - kinetic friction
= 100 - 98.1 * 0.4 = 61 N
=> 10 kg block will slide on 40 kg slab with 61/10 = 6.1 m/s

Frictional force on 40 kg slab by 10 kg block = 98.1 * 0.4 = 39 N
=> 40 kg slab will move with
39/40 m/s²
= 0.98 m/s²

Answer 3

Remember shortcut.... In this kind of case... Acceleration=(coefficient of friction*m1*g)/m2.....
So. a=(.4*10*10)/40=1m/s²

Answer 4

Given m =10 kg
And we have F = Uķmg
=0.40×10 ×10
=40N
And, F= ma
a=F/m= 40/40= 1 m/s^2

Answer 5

Bad worst

Answer 6

mass of block = 10Kg
F = (coefficient of friction)* mg
=0.40*10*10
=40N
By F = ma
a = F/m = 40/40 = 1 m/s^2