11. A block of mass 200 g is moving with a velocity of 5 m/
sec along the positive x-direction. At time 1 = 0, when the
body is at x = 0, a constant force 0.4 N is directed along the
negative x-direction, is applied on the body for 10 s. What
is the position x of the body at T = 2.5 s?
(1) x= 1.75 m
(2) X= 1.25 m
(3) x= 1.0 m
(4) x= 1.5 m
et
Answers
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The position x of the body at T = 2.5 s is 6.25 m.
Given: A block of mass 200 g is moving with a velocity of 5 m/sec along the positive x-direction. At time 1 = 0, when the body is at x = 0, a constant force 0.4 N is directed along the negative x-direction, which is applied to the body for 10 s.
To Find: The position x of the body at T = 2.5 s.
Solution:
- We know that we can relate force and acceleration using the formula,
F = m × a ....(1)
Where F = force, m = mass, a = acceleration.
- We know that we can solve this using the formula of laws of motion,
S = ut + 1/2 × at² ....(2)
Where S = distance, u = initial velocity, a = acceleration, t = time.
Coming to the numerical, we are given;
The initial velocity (u) = 5 m/s
Mass of the block (m) = 200 g = 0.2 kg
The magnitude of constant force = - 0.4 N [minus for negative direction]
Time for which distance is to be calculated (t) = 2.5 s
Putting respective values in (1), we get;
F = m × a
⇒ - 0.4 = 0.2 × a
⇒ a = - 2 m/s²
Now, we need to find the position of the body at t = 2.5 s.
So, putting respective values in (2), we get;
S = ut + 1/2 × at²
⇒ S = ( 5 × 2.5 ) + 1/2 × ( -2 ) × ( 2.5 )²
⇒ S = 12.5 - 6.25
⇒ S = 6.25 m
Hence, the position x of the body at T = 2.5 s is 6.25 m.
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