Answer:
(a) 1.2 J
(b) 5 m/s
(c) 6.3 J
Explanation:
(a) Kinetic energy at A
Ek = 1/2mv².................. Equation 1
Where Ek = Kinetic energy at A, m = mass of the particle, v = velocity at A.
Given: m = 0.6 kg, v = 2.0 m/s
Substitute into equation 1
Ek = 1/2(0.6)(2²)
Ek = 0.6(2)
Ek = 1.2 J.
(b)
Speed at point B
Ek' = 1/2mv'²............... Equation 2
Make v' the subject of the equation,
v' = √(2Ek'/m).................. Equation 3
Where, Ek' = kinetic energy at B, v' = velocity at B.
Given: Ek' = 7.5 J, m = 0.6 kg,
Substitute into equation 3
v' = √[(2×7.5)/0.6]
v' =√(15/0.6)
v' = √25
v' = 5 m/s.
(c)
Wt = Δ kinetic energy from A to B
Where Wt = total work done as the particle moves from A to B.
Wt = 1/2m(v'²-v²)
Wt = 1/2(0.6)(5²-2²)
Wt = 0.3(25-4)
Wt = 0.3(21)
Wt = 6.3 J
