Laws of Motion
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Question : 10
Total: 40
A body of mass 0.40 kg moving initially with a constant speed of10 m s– 1
Solution:
Here, m = 0.40 kg, u = 10 m s–1 due north
F = – 8.0 N (minus sign for opposite direction of force)
a =
=
= − 20 m s − 2 0 ≤ t ≤ 30 s
(i) At t = – 5 s, x = ut
= 10 × (–5) = – 50 m
(ii) Att = 25 s ,
x = u t +
a t 2
= 10 × 25 +
( − 20 ) ( 25 ) 2
= − 6000 m
(iii) At t = 100 s,
The problem is divided into two parts. First consider motion up to 30 s
x 1 = u t +
a t 2
= 10 × 30 +
( − 20 ) ( 30 ) 2
= − 8700 m
At t = 30 s, v = u + at
= 10 – 20 × 30 = – 590 m s– 1
For motion 30 s to 100 s
x 2 = v t
= − 590 × 70 = − 41300
∴ x = x 1 + x 2
= – 8700 − 41300
= – 50000 m = – 50 k m
F = – 8.0 N (minus sign for opposite direction of force)
(i) At t = – 5 s, x = ut
= 10 × (–5) = – 50 m
(ii) At
(iii) At t = 100 s,
The problem is divided into two parts. First consider motion up to 30 s
At t = 30 s, v = u + at
= 10 – 20 × 30 = – 590 m s
For motion 30 s to 100 s
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