Question

Wedge of 10 kg is free to move on horizontal surface. At the given
instant, acceleration of wedge is (string and pulleys are ideal)
(A) 2 m/s2 towards right
(B) 2 m/s2 towards left
(C) 1 m/s2 toward left
(D) 1 m/s2 toward right​

Answer 1

Answer:

(D) 1 m/s^2 towards right

Explanation:

1) Let the tension in the string be 'T'.

Since, Tension is constant all along the string .

=> F=T=50 N.

2) We know,

Wedge of 10 kg is free to move, that is, it will move on horizontal plane.

Net Force on wedge on right, = F cos(37)

                                           = 50 *4/5 = 40 N

Net Force on wedge on left, =  T cos(53)

                                                = 50*3/5=30 N

Acceleration of wedge is given by ,

$a=\frac{F_{net}}{M} =\frac{40-30}{10}\\ \\ =\frac{10}{10}=1 m/s^2$

Since, Right force is greater than left.

Hence, Acceleration of wedge is 1 m/s^2 towards right .

Answer 2

Answer:

(D) 1 m/s^2 towards right

Explanation:

1) Let the tension in the string be 'T'.

Since, Tension is constant all along the string .

=> F=T=50 N.

2) We know,

Wedge of 10 kg is free to move, that is, it will move on horizontal plane.

Net Force on wedge on right, = F cos(37)

= 50 *4/5 = 40 N

Net Force on wedge on left, = T cos(53)

= 50*3/5=30 N

Acceleration of wedge is given by ,

a=\frac{F_{net}}{M} =\frac{40-30}{10}\\ \\ =\frac{10}{10}=1 m/s^2

Since, Right force is greater than left.

Hence, Acceleration of wedge is 1 m/s^2 towards right .