A 5.0 kg block of ice is at rest at the top of a smooth inclined plane. The block is released and slides 2.0 m down the plane. Assuming there is no friction between the block and the surface, calculate a) the gravitational potential energy at the top of the plane
b) the component of the weight parallel to the plane
c) the acceleration of the block
d) the velocity of the block at the bottom of the plane
e) the kinetic energy at the bottom of the plane. ​

a) 98.1 Joules

b) 49.05 N × sin(θ)

c) 9.81 × sin(θ)

d) The velocity of the block at the bottom of the plane, v is approximately 6.264 m/s

e) 98.1 Joules

Explanation:

The given parameters of the block are;

The mass of the block, m = 5.0 kg

The distance down the plane the block slides, h = 2.0 m

The friction between the block and the surface = 0

Let θ represent the angle of inclination oof the plane

a) The gravitational potential energy, P.E. = m·g·h

Where;

g = The acceleration due to gravity ≈ 9.81 m/s²

∴ P.E. ≈ 5.0 kg × 9.81 m/s² × 2.0 m = 98.1 Joules

The gravitational potential energy, P.E. ≈ 98.1 Joules

b) The component of the weight of the block parallel to the plane, , is given as follows;

= w × sin(θ) = m·g·sin(θ)

∴ ≈ 5.0 kg × 9.81 m/s² × sin(θ) = 49.05 × sin(θ) N

The component of the weight of the block parallel to the plane, ≈ 49.05 N × sin(θ)

c) The component of the weight along the inclined plane = The force with which the block moves along the inclined plane, therefore;

= m·g·sin(θ) = m·a

Where a represents the acceleration of the block along the plane

Therefore, by comparison, we have;

g·sin(θ) = a

∴ a ≈ 9.81 × sin(θ)

d) Given that the motion of the block is 2.0 m downwards, we have;

The velocity of the block at the bottom of the plane, v² = 2·g·h

Therefore, v² ≈ 2 × 9.81 m/s²× 2.0 m = 39.24 m²/s²

v = √(39.24 m²/s²) ≈ 6.264 m/s

e) The kinetic energy at the bottom of the plane, K.E. = (1/2)·m·v²

∴ K.E. = (1/2) × 5.0 kg × 39.24 m²/s² = 98.1 J

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hope this