Question

When the system shown here is pulled by force F, the 1 kg block remains stationary relative to the wedg
F
1kg
3kg
37°
The force Fis 7.5 N
The normal force exerted by the wedge on the block is 8 N
15
The acceleration of the system is
m/s2
8
The tension in string is 6.25 N​

Answer 1

Explanation:

From figure, there is no acceleration in vertical direction. So, Applying Newton's Law in vertical direction, we get

N=mg

=50N

Maximum value of Friction f

max

=μN=0.2×50=10 N

Since, f

max

≤F

So, sliding will happen, and kinetic friction will act, i.e.

f

kinetic

=μN=10N

Step 3: Newton’s Law for finding acceleration

Applying Newton's second law on block, in horizontal direction

(Taking rightward positive)

∑F=ma

⇒F−f

kinetic

=ma

⇒F−μmg=ma

⇒40−0.2×5×10=5a

⇒a=6m/s

2

Hence acceleration of the block will be 6m/s

2

Hence, Option A is correct.

solution