Force acts for 20 s on a body of mass 20 kg, starting from rest, after which the force ceases and then body describes 50 m in the next 10 s. The value of force will be :

Question

Force acts for 20 s on a body of mass 20 kg, starting from rest, after which the force ceases and then body describes 50 m in the next 10 s. The value of force will be : (A) 40 N (B) 5 N (C) 20 N (D) 10 N

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/aa545428e69a2b513020471ebdaaf5a2-.png" /> 50=V × 10 V=5 m / s V =0+ a × 20 5= a × 20 a =(1/4) m / s 2 F = ma =20 × (1/4)=5 N

Q. Force acts for $20 s$ on a body of mass $20 k g$ , starting from rest, after which the force ceases and then body describes $50 m$ in the next $10 s$ . The value of force will be :

40 N

5 N

20 N

10 N

$50 = V \times 10$
$V = 5 m / s$
$V = 0 + a \times 20$
$5 = a \times 20$
$a = \frac{1}{4} m / s^{2}$
$F = ma = 20 \times \frac{1}{4} = 5 N$

Q. Force acts for 20s on a body of mass 20kg, starting from rest, after which the force ceases and then body describes 50m in the next 10s. The value of force will be :

40 N

5 N

20 N

10 N

50=V×10 V=5m/s V=0+a×20 5=a×20 a=41​m/s2 F=ma=20×41​=5N

Q. Force acts for $20 s$ on a body of mass $20 k g$ , starting from rest, after which the force ceases and then body describes $50 m$ in the next $10 s$ . The value of force will be :

$50 = V \times 10$
$V = 5 m / s$
$V = 0 + a \times 20$
$5 = a \times 20$
$a = \frac{1}{4} m / s^{2}$
$F = ma = 20 \times \frac{1}{4} = 5 N$