A force of 3^i−15^j acts on 5 kg body The body is at a
|
A force of (3^i−1.5^j) acts on 5 kg body. The body is at a position of (2^i−3^j)m and is travelling at 4ms−1.
The force acts on the body until it is at the position (^i+5^j) m. Assuming no other force does work on the body, the final speed of the body is
A.
√10m/s
B.
5 m/s
C.
√5m/s
D.
10 m/s
Please scroll down to see the correct answer and solution guide.
Right Answer is: A
SOLUTION
Given, mass of the body = 5 kg
Force →F=3^i−1.5^j
∴→a=32^i−310^j
Now displacement , →s={(^i+5^j)−(2^i−3^j)}m=(−^i+8^j)m
From work energy principle
W=→F.→S=12m(v2−u2)
⇒v=√10m/s