Question

BALOK ON BLAOK
104. A 40 kg slab rests on a frictionless floor. A
10 kg block rests on top of the slab. The
static coefficient of friction between the
block and slab is 0.60 while the kinetic
coefficient is 0.40. The 10 kg block is acted
upon by a horizontal force of 100N. Ifg=
9.8 m/s 2, the resulting acceleration of the
slab will be :-
No Fnction
100N
10kg
40g
2) 1.47 m/s
1) 0.98 m/s2
3) 1.52 m/s
105. In the
40 6.1 m/s​

Answer 1

Answer:

0.98 m/s 2

Explanation:

total mass (40+10) kg first. 

Using formula F=ma or a=mF=50kg100N.

The acceleration is 2m/s2  and the maximum frictional force =μs×N

=μs×m1g

=(0.60)(10)(9.8)=58.8N

Thus, we see frictional force is smaller than the applied force and proves that the block and the slab do not move relative to each other.

For the block we have F=μmg

=0.4×10×9.8=39.2N

Resulting acceleration of the slab , a=F/m=39.2/40=0.98m/s2

Thus, the resulting acceleration of the slab is 0.98m/s2

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