here,
mass , m = 2 kg
at x1 = 1 m and v1 = 3 m/s
a)
the total work done on the crate along the floor , Wt = area under the curve
Wt = (4 * 1) + ( -4) * ( 2 - 1 ) + 0.5 * ( 3 - 2) * (-4)
Wt = - 2 J
b)
let the final velocity at x = 3 m be v
using work energy theorm
work done between x=1 m to x=3 m = kinetic energy gained
( -4) * ( 2 - 1 ) + 0.5 * ( 3 - 2) * (-4) = 0.5 * 2 * (v^2 - 3^2)
solving for v
v = 1.73 m/s
the speed at x=3 m is 1.73 m/s
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