A constant force acts on a body of mass 0.9 kg at rest for 10 s. If the body moves a distance of 250 m the magnitude of the force is:

Question

A constant force acts on a body of mass 0.9 kg at rest for 10 s. If the body moves a distance of 250 m the magnitude of the force is: (A) 3 N (B) 3.5 N (C) 4.0 N (D) 4.5 N

Tardigrade Admin · Accepted Answer

By using the equation S = ut +(1/2) a t 2 As the initial velocity is zero By substituting the values we get 250=50 a The acceleration is 5 ( m / sec 2) We know that F = m × a We get F =0.9 × 5 Hence the force is 4.5 N

Q. A constant force acts on a body of mass $0.9 k g$ at rest for $10 s$ . If the body moves a distance of $250 m$ the magnitude of the force is:

3 N

3.5 N

4.0 N

4.5 N

By using the equation
$S = u t + \frac{1}{2} a t^{2}$
As the initial velocity is zero
By substituting the values we get
$250 = 50 a$
The acceleration is $5 \frac{m}{s e c ^{2}}$
We know that
$F = m \times a$
We get
$F = 0.9 \times 5$
Hence the force is $4.5 N$

Q. A constant force acts on a body of mass 0.9kg at rest for 10s. If the body moves a distance of 250m the magnitude of the force is:

3 N

3.5 N

4.0 N

4.5 N

By using the equation S=ut+21​at2 As the initial velocity is zero By substituting the values we get 250=50a The acceleration is 5sec2m​ We know that F=m×a We get F=0.9×5 Hence the force is 4.5N

Q. A constant force acts on a body of mass $0.9 k g$ at rest for $10 s$ . If the body moves a distance of $250 m$ the magnitude of the force is:

By using the equation
$S = u t + \frac{1}{2} a t^{2}$
As the initial velocity is zero
By substituting the values we get
$250 = 50 a$
The acceleration is $5 \frac{m}{s e c ^{2}}$
We know that
$F = m \times a$
We get
$F = 0.9 \times 5$
Hence the force is $4.5 N$