Question

class 11th laws of motion​

Answer 1

⟶ AnSwer :-

The free body diagram of the man and the platform can be drawn as shown in the figure.

[Note :- If the tension in the string is equal to the force by which he pulls the rope]

(i) Applying Newton's second law

2T- (M + m)g = (M + m)a

$\sf a = \dfrac{2T- (M + m)g }{M + m}$

Here,

• M = 60 kg
• m = 15 kg
• T = 400 N
• g = 10 m/s²

$\sf a = \dfrac{2(400) - (60 + 15)(10)}{60 + 15} = 0.67 {ms}^{ - 2}$

Thus, the acceleration of the platform is 0.67m/s²

(ii) To attain a speed of m/s in one second, the acceleration a must be 1m/s²

Then, 2T = (M + m)(g + a)

$\sf T= \dfrac{(60 + 15)(10 + 1)}{2} = 412.5 \: N$

thus, He must exert 412.5 N on the rope to attain an upward speed of 1m/s in 1s