Answer:
a) 98.1 Joules
b) 49.05 N à sin(θ)
c) 9.81 à 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 â 9.81 m/s²
â´ P.E. â 5.0 kg Ă 9.81 m/s² Ă 2.0 m = 98.1 Joules
The gravitational potential energy, P.E. â 98.1 Joules
b) The component of the weight of the block parallel to the plane, [tex]w_{\parallel}[/tex], is given as follows;
[tex]w_{\parallel}[/tex] = w à sin(θ) = m¡g¡sin(θ)
â´ [tex]w_{\parallel}[/tex] â 5.0 kg Ă 9.81 m/s² Ă sin(θ) = 49.05 Ă sin(θ) N
The component of the weight of the block parallel to the plane, [tex]w_{\parallel}[/tex] â 49.05 N Ă sin(θ)
c) The component of the weight along the inclined plane = The force with which the block moves along the inclined plane, therefore;
[tex]w_{\parallel}[/tex] = 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
â´ a â 9.81 Ă 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 â 2 Ă 9.81 m/s²à 2.0 m = 39.24 m²/s²
v = â(39.24 m²/s²) â 6.264 m/s
e) The kinetic energy at the bottom of the plane, K.E. = (1/2)¡m¡v²
ⴠK.E. = (1/2) à 5.0 kg à 39.24 m²/s² = 98.1 J
