Question : A body of mass 0.5 kg travels in a straight line with velocity v=ax3/2 where a=5m-1/2s-1. The work done by the net force during its displacement from x =0 to x = 2 m is
a) 1.5 J
b) 50 J
c) 10 J
d) 100 J
Doubt by Muskan
Ans : b) is the correct option.
Detailed Solution :
Method I
Method II
m = 0.5 kg
v=ax3/2 where a=5m-1/2s-1
Velocity at x=0
v=a(0)3/2
v=0 m/s
v=0 m/s
So, initial velocity (u) = 0 m/s
Velocity at x=2
v=a(2)3/2
v=a(2√2)
v=(5)(2√2)
v=10√2 m/s
v=a(2√2)
v=(5)(2√2)
v=10√2 m/s
So, final velocity (v) = 10√2 m/s
We know,
As per the Work Energy Theorem
Work Done = Change in Kinetic Energy
As per the Work Energy Theorem
Work Done = Change in Kinetic Energy
W = ½mv2 - ½mu2
W = ½m(v2-u2)
W = ½(0.5)×[(10√2)2-(0)2]
W = ½(0.5)×[200-0]
W = ½×0.5×200
W = 100/2
W = 50 J
W = ½m(v2-u2)
W = ½(0.5)×[(10√2)2-(0)2]
W = ½(0.5)×[200-0]
W = ½×0.5×200
W = 100/2
W = 50 J
Hence, b) would be the correct option.