Answer:
Explanation:
Work done by force applied = force x displacement
= 10 x 2 = 20 J
Negative work done by frictional force
= μ mg x d where μ is coefficient of kinetic friction , m is mass and d is displacement
= - .18 x 4.5 x 9.8 x 2
= - 15.87 J
Net positive work done on the mass = 4.13 J
If v was the initial velocity
increase in kinetic energy = positive work done on mass
= 1 / 2 m v² - 1/2 m u² = 4.13 where v is final and u is initial velocity
1 /2 x 4.5 x 2 ² - 1/2 x 4.5 u² = 4.13
9 - 2.25 u² = 4.13
2.25 u² = 4.87
u² = 2.16
u = 1.47 m /s .
