Question : A body of mass 0.5 kg travels in straight line with velocity V=ax3/2 where a=5 m-1/2s-1. The work done by the net force during it's displacement from x=0 to x=2m is (a) 15 J (b) 50 J (c) 10 J (d) 100 J
Doubt by Sanskriti
Solution :
m=0.5 kg
V=ax3/2
a=5 m-1/2s-1
So
V=5x3/2
V=5x3/2
As per the work energy therem
W=Final KE - Initial KE
Initial KE at x=0
K1=½mV²
K1=½m[5x^3/2]².
W=Final KE - Initial KE
Initial KE at x=0
K1=½mV²
K1=½m[5x^3/2]².
K1=½m[5(0)3/2]²
K1=0
K1=0
Final KE at x=2m
K2=½mV²
K2=½m[5x3/2]².
K2=½m[5x3/2]².
K2=½m[5(2)3/2]²
K2=½m[5×√8]²
K2=½m[5×√8]²
K2=½ m×25×8
K2=100 m
K2=100×0.5
K2=100 m
K2=100×0.5
K2=50 J
W=K2-K1
W=50-0
W=K2-K1
W=50-0
W=50 J
Hence, (b) 50 J, would be the correct option.