8
the distance X moved by a body of mass 0.5 kg
under the action of a force varies with time t as
vm) = 3t²+ 4t + 5 (here t is expresed in second)
What is the work done by the force in first
2 seconds?
(2) 40 J
(3) 50 J
(4) 60 J
(1) 20 J
Answers
Answer:
Work done = 40 J
Explanation:
Given ;
Mass ( m ) = 0.5 kg
putting t = 2 sec we get
a = 6 × 2 + 4
Now from Newton second law of motion
Here u = 0
s = 1 / 2 × 16 × 2 × 2
s = 32 m
Now for work done ;
W = 0.5 × 16 + 32 J
W = 8 + 32 J
W = 40 J
Thus work done is 40 J
Correct question :-
The distance x moved by a body of mass 0.5 kg by a force varies with time t as x = 3t² + 4t + 5, where x is expressed in metres and t in seconds. What is the work done by the force in the first 2 seconds?
Answer :-
Work done = 75 J
Explanation :-
Mass = 0.5 kg
distance = x = 3t² + 4t + 5
_____________________[Given]
We know that,
Acceleration (a) = d²(x)/dt²
=> a = d²(3t² + 4t + 5)/dt²
=> a = 6 m/s²
______________________________
Also,
Force (F) = mass × acceleration
=> F = 0.5 × 6
=> F = 3 N
_______________________________
Now,
Distance travelled = x = 3t² + 4t + 5
Put t = 2 sec , we get
x = 3(2)² + 4(2) + 5
=> x = 12 + 8 + 5
=> x = 25 m
_______________________________
Now,
To find :- Work done = ?
We have,
Work done (W) = Force × distance
=> W = 3 × 25
=> W = 75 J
______________________________
Hence, the work done is 75 J
________________________[Answer]