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1. A body of mass 500 g, initially at rest, is acted upon
by a force which causes it to move a distance of 4
m in 2 s. Calculate the force applied.
Answers
- Mass of body (m) = 500g or 0.5kg
- Initially body is at rest (u) = 0
- When force is applied it moves by distance (s) = 4m in time (t) =2s
- Force applied ( F )
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❏ Velocity
v = finial Velocity of body
s = distance
t = time
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❏ Acceleration
From Newton's First Law Of Motion
v = final velocity of body
u = Initial velocity of body
a = acceleration
t = time
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From Newton's Second Law
F = force applied
m = mass of body
a = acceleration
Force applied to the body is
Answer:
- The Force (F) Applied on the body is 1 N.
Given:
- Mass of the body (M) = 500 g = 0.5 Kg
- Distance covered (S) = 4 m
- Time taken (t) = 2 sec.
- Initial velocity (u) = 0 m/s.
Explanation:
From second Kinematic equation we know,
⇒ S = u t + 1 / 2 a t²
Where,
- S Denotes Distance travelled.
- u Denotes Initial velocity.
- t Denotes Time taken.
- a Denotes Acceleration.
Now,
⇒ S = u t + 1 / 2 a t²
Substituting the values,
⇒ 4 = 0 × 2 + 1 / 2 × a × (2)²
⇒ 4 = 0 + 1 / 2 × a × 4
⇒ 4 = 1 / 2 × a × 4
⇒ 4 = a × 2
⇒ a = 4 / 2
⇒ a = 2
⇒ a = 2 m/s²
∴ We got the acceleration.
From Newton's Second law
⇒ F = M a
Where,
- F Denotes Force.
- M Denotes Mass.
- a Denotes acceleration.
Now,
⇒ F = M a
Substituting the values,
⇒ F = 0.5 Kg × 2 m/s²
⇒ F = 0.5 × 0.2
⇒ F = 1
⇒ F = 1 N.
∴ The Force (F) Applied on the body is 1 N.