Question

A 10 kg block is kept on an incline plane pulled by a string applying 200 newton force of 10 newton force is

Answer 1

Answer:

The answer is 200√ 2 N.

Explanation:

Tension in the string "T" = 200 N

Let acceleration be "a"

200 - 10 - 100sin30 = 10a

a = 140 / 10

a = 14 m/s^2

Net force on pulley = √  (200)^2 + (200)^2

                                = 200√ 2 N

Thus the tension in the string is 200√ 2 N.

Answer 2

Acceleration- 14m/s²

Force Acting- 200√2N

(i) Net force at 30° inclination;

applied force: 10N

weight component: (10g)(sin30) = 50N

net force; 10N+50N = 60N

(ii) At 60° inclination force is applied of 200N

Since this force is greater than 60N force. Therefore block will accelerate in the direction of 200N

(iii) According to FBD (as shown in image);

(200)-(60)=(10)(a)

(140)(a)=(10a)→ (a=14m/s²)

(iv) Net force on pulley;

F = √(200)²+(200)² =20√2N

Note: We can ignore the normal force of block acting as it is a internal force canceled by normal force of wedge. Also please see the attached image of FBD.

Hope It Helps!