Question

block of mass 5 kg is present on a plank of mass 10 kg. The coefficient of friction
between them is 0.8 and that between plank and the ground is 0.4. If the plank is
being pulled with a constant force 90 N to the right, find their accelerations​

Answer 1

Answer:

For 10kg block, f is frictional force

f

1max

=μ

1

μ

2

=μ

1

(10g)=0.1(10×10)=10N

i) if F=2N then acting friction

f

1

=F=2N

(A) is correct.

for 5 kg block;

f

2max

=μ(10+5)g=0.3×15×10=45N

ii) if F=30N then;

F−F

1max

=10a

10a=30−10=20

a=2m/s

2

(B) is correct.

iii) if μ

1

changed to 0.5;

then f

1max

=μ

1

(10g)=0.5×10×10=50N

In the case acting friction will be F

2max

Since,

F

1max

=F

2max

Block 10kg and 5kg both move together then F

min

to begin is

F

min

−F

max

=0

F

min

=F

2max

=45N

(c) is correct.

iv) if μ=0.1 , then f

max

=10N and,

f

2max

=45N

f

2max

>f

1max

always for any any value of N. Hence, 5kg block never move on the ground.

So, option (A), (B), (C), and (D) are correct.

Answer 2

Given:

Mass = 5 kg

Plank mass = 10 kg

Friction between block and plank = $0.8$

Friction between plank and ground = $0.4$

Force = 90 N

To find:

We need to find the acceleration.

Solution:

Acceleration is that the rate of amendment of rate. Usually, acceleration suggests that the speed is dynamic, however not continuously. once an object moves in an exceedingly circular path at a continuing speed, it's still fast, as a result of the direction of its rate is dynamic.

$& F=ma \\ \\ & a=\frac{F}{m}$

So,

$a=\frac{90}{15}=6\text{ m}{{\text{s}}^{2}}$

Maximum possible acceleration of 5kg block.

$& {{a}_{1}}=\frac{FrictionForce}{mass} \\ & \\ & {{a}_{1}}=\frac{90}{5} \\ & \\ & {{a}_{1}}=18\text{ }m{{s}^{2}}$

Therefore, the acceleration is $\[6m{{s}^{2}}\]$ and $\[18m{{s}^{2}}\]$.