A 2 kg block slides down a frictionless incline making an angle of 51 with the h

Question

A 2 kg block slides down a frictionless incline
making an angle of 51 with the horizontal.
The acceleration of gravity is 9.81 m/s2 .

a) Find the work done by the gravitational
force when the block slides 4 m (measured
along the incline).
Answer in units of J

b) What is the total work done on the block?
Answer in units of J

c) What is the speed of the block after it has
moved 4 m if it starts from rest?
Answer in units of m/s

d) What is its speed after 4 m if it starts with an initial speed of 1.8 m/s?
Answer in units of m/s

Explanation / Answer

M = 2 kg ,

Weight of the block = W = Mg = 2*(9.81) N
Angle made by the inclined plane with the horizontal = = 51 deg
Resolve the weight in to 2 components :
(1) Perpendicular or normal to the plane, W cos = Mg cos = 2*(9.81)*(0.63) N
which is balanced by the Normal reaction of the plane, R
Normal reaction of the plane = R = Mg cos
(2) Parallel to the inclined plane, W sin = Mg sin = 2*(9.81)*(0.78) = 15.28 N
which is unbalanced

(a) Distance moved along the incline = l = 4 m
Work done by the gravitational force = (Mg sin ) l = Mg l sin = 2*(9.81)*(0.78)*(4) J = 61.21 J

(b) Total work done on the block = Mg l sin = 2*(9.81)*(0.78)*(4) = 61.21 J

(c) Distance moved = l = 3.9 m , Initial speed = 0, Acceleration = a = Mg sin / M = g sin
Final speed = v ; v² = u² + 2 a l = 0 + 2 (g sin ) l = 2*(9.81)*(0.78)*(4) = 61.21
v = (61.21) = 7.82 m/s

(d) Initial speed = u = 1.8 m/s , Acceleration = a = g sin , Distance moved = l = 3.9 m
Final speed = V
V² = u² + 2 a l = (1.8)² + 2 (9.81)*(0.78)*(4) = 3.24 + 61.21 = 64.45
V = (64.45) = 8.02 m/s

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